Electric Charges And FieldsClass 12 Physics NCERT Solutions

The force between two small charged spheres can be calculated using Coulomb's Law.

Coulomb's Law: The magnitude of the electrostatic force $F$ between two point charges is given by: $F = k \frac{|q_{1} q_{2}|}{r^{2}}$

Calculation: Substituting the given values into the formula: $F = (9 \times 10^{9}) \times \frac{|(2 \times 10^{-7}) \times (3 \times 10^{-7})|}{(0.3)^{2}}$ $F = (9 \times 10^{9}) \times \frac{6 \times 10^{-14}}{0.09}$ $F = (9 \times 10^{9}) \times \frac{6 \times 10^{-14}}{9 \times 10^{-2}}$ $F = 6 \times 10^{9 - 14 + 2}$ $F = 6 \times 10^{-3} \mathrm{~N}$

Since both charges are positive, the force is repulsive in nature.

Answer: The force between the two small charged spheres is $6 \times 10^{-3} \mathrm{~N}$ and it is repulsive.

This problem involves applying Coulomb's Law and Newton's third law of motion.

Coulomb's Law: The magnitude of the force is given by: $F = k \frac{|q_{1} q_{2}|}{r^{2}}$ We need to find the distance $r$. Rearranging the formula: $r^{2} = k \frac{|q_{1} q_{2}|}{F}$

Calculation: Substituting the given values: $r^{2} = (9 \times 10^{9}) \times \frac{|(0.4 \times 10^{-6}) \times (-0.8 \times 10^{-6})|}{0.2}$ $r^{2} = (9 \times 10^{9}) \times \frac{0.32 \times 10^{-12}}{0.2}$ $r^{2} = (9 \times 10^{9}) \times (1.6 \times 10^{-12})$ $r^{2} = 14.4 \times 10^{-3} = 144 \times 10^{-4} \mathrm{~m}^{2}$ $r = \sqrt{144 \times 10^{-4}} = 12 \times 10^{-2} \mathrm{~m} = 0.12 \mathrm{~m}$ So, the distance between the two spheres is 12 cm.

According to Newton's third law of motion, for every action, there is an equal and opposite reaction. The electrostatic force is a mutual force. Therefore, the force exerted by the first sphere on the second sphere ($F_{21}$) is equal in magnitude and opposite in direction to the force exerted by the second sphere on the first ($F_{12}$).

$\mathbf{F}_{21} = -\mathbf{F}_{12}$ The magnitude of the force is the same. $|F_{21}| = |F_{12}| = 0.2 \mathrm{~N}$ Since the charges are opposite in sign (one positive, one negative), the force is attractive.

Answer: (a) The distance between the two spheres is 0.12 m or 12 cm. (b) The force on the second sphere due to the first is 0.2 N, and it is attractive.

To check if the ratio is dimensionless, we will analyze the units of each constant in the expression.

Now, let us find the units of the numerator and the denominator.

Since the units of the numerator and the denominator are the same, their ratio will be a unitless, and therefore dimensionless, quantity.

Using the standard values of the physical constants:

Now, we calculate the ratio: $\frac{k e^{2}}{G m_{e} m_{p}} = \frac{(8.987 \times 10^{9}) \times (1.602 \times 10^{-19})^{2}}{(6.674 \times 10^{-11}) \times (9.109 \times 10^{-31}) \times (1.672 \times 10^{-27})}$ $\frac{k e^{2}}{G m_{e} m_{p}} = \frac{2.304 \times 10^{-28}}{1.015 \times 10^{-67}} \approx 2.27 \times 10^{39}$

The expression $k e^{2}$ is related to the electrostatic force ($F_e$) between an electron and a proton, and the expression $G m_{e} m_{p}$ is related to the gravitational force ($F_g$) between them.

The ratio we calculated is therefore the ratio of the magnitudes of the electrostatic force to the gravitational force between an electron and a proton at any given distance $r$. $\frac{F_e}{F_g} = \frac{k e^{2} / r^2}{G m_{e} m_{p} / r^2} = \frac{k e^{2}}{G m_{e} m_{p}} \approx 2.27 \times 10^{39}$

Conclusion: The ratio signifies that the electrostatic force between an electron and a proton is about $10^{39}$ times stronger than the gravitational force between them. This shows why electrostatic forces dominate at the atomic and molecular level, while gravitational forces are negligible in that domain.

The statement 'electric charge of a body is quantised' means that electric charge is not a continuous quantity but occurs in discrete packets or units. All free charges are integral multiples of a basic, fundamental unit of charge, denoted by 'e'. The value of this basic unit of charge is the magnitude of the charge on a single electron or proton, which is approximately $1.602 \times 10^{-19}$ Coulombs.

Mathematically, the total charge (q) on any body can be expressed as: $q = ne$ where 'n' is an integer (n = 0, ±1, ±2, ±3, ...). This implies that a body cannot have a charge of, for example, 0.5e or 2.75e. Charge can only be added to or removed from a body in steps of 'e'.

When dealing with macroscopic or large-scale charges, the quantisation of electric charge can be ignored because the basic unit of charge, 'e', is extremely small.

Macroscopic charges, such as those encountered in everyday life (e.g., 1 microcoulomb, $1 \mu \mathrm{C}$), consist of an enormous number of basic charge units. For example, a charge of $1 \mu \mathrm{C}$ contains approximately $6.25 \times 10^{12}$ electrons.

Since the number of elementary charges is so large, the addition or removal of a few electrons does not cause a noticeable change in the total charge. The discrete nature of charge is not apparent, and the charge distribution appears to be continuous. This is analogous to viewing a dotted line from a great distance; it appears as a continuous line because the individual dots are too small and close together to be resolved. Therefore, for practical purposes at the macroscopic level, we can treat charge as a continuous quantity.

The law of conservation of charge states that the total electric charge in an isolated system remains constant. Charge can neither be created nor destroyed; it can only be transferred from one body to another.

The observation of charges appearing on a glass rod and a silk cloth when they are rubbed together is perfectly consistent with this law. Here is the explanation:

Thus, the phenomenon of charging by friction demonstrates a transfer of charge, not its creation or destruction, which is in perfect agreement with the law of conservation of charge.

Let us analyze the forces exerted on the $1 \mu \mathrm{C}$ charge (let us call it $q_O$) placed at the center O of the square ABCD.

The center O is equidistant from all four corners. The distance from the center to any corner is half the length of the diagonal. Diagonal $d = \sqrt{s^2 + s^2} = s\sqrt{2} = 10\sqrt{2}$ cm. Distance from center to a corner, $r = d/2 = 5\sqrt{2}$ cm. So, OA = OB = OC = OD = $5\sqrt{2}$ cm.

We will use the principle of superposition to find the net force on the charge $q_O$. We calculate the force exerted by each corner charge on $q_O$ and then find their vector sum.

Net Force: The total force on the charge $q_O$ is the vector sum of all the individual forces: $\mathbf{F}_{net} = \mathbf{F}_{OA} + \mathbf{F}_{OB} + \mathbf{F}_{OC} + \mathbf{F}_{OD}$ $\mathbf{F}_{net} = (\mathbf{F}_{OA} + \mathbf{F}_{OC}) + (\mathbf{F}_{OB} + \mathbf{F}_{OD})$ $\mathbf{F}_{net} = 0 + 0 = 0$

Answer: The net force on the charge of $1 \mu \mathrm{C}$ placed at the centre of the square is zero.

An electrostatic field line represents the path that a small positive test charge would take if it were free to move in the electric field. The tangent to the field line at any point gives the direction of the electric force (and hence the electric field) at that point.

If a field line had a sudden break, it would imply that the electric field abruptly becomes zero at the point of the break. However, the electric field due to a charge distribution is continuous in a charge-free region. A moving test charge experiences a continuous force, so its path, which is represented by the field line, must also be continuous. A break in the line would mean there is no field, and thus no force, in that gap, which is physically impossible unless the source charges are arranged in a very specific way to make the field zero at that point, but then the field line would terminate there, not break and restart.

Therefore, an electrostatic field line must be a continuous curve.

By definition, the tangent to an electric field line at any point gives the direction of the net electric field at that point. The electric field at any point in space is a vector quantity and has a unique direction.

If two field lines were to cross each other at a point, it would mean that at the point of intersection, there would be two different tangents. This would imply that the electric field at that single point has two different directions simultaneously. This is physically impossible, as the net force on a test charge at a single point can only be in one specific direction.

Therefore, two electric field lines can never intersect or cross each other.

This problem involves an electric dipole configuration.

The net electric field at O is the vector sum of the electric field due to $q_A$ and the electric field due to $q_B$.

The force $\mathbf{F}$ on a charge $q$ in an electric field $\mathbf{E}$ is given by $\mathbf{F} = q\mathbf{E}$.

Calculation: Magnitude of the force: $F = |q_{test}| \times E_{net} = (1.5 \times 10^{-9}) \times (5.4 \times 10^6) = 8.1 \times 10^{-3} \mathrm{~N}$

Direction of the force: Since the test charge is negative, the force on it will be in the direction opposite to the electric field. The electric field is directed from A to B. Therefore, the force on the negative test charge will be directed from B to A.

Answer: (a) The electric field at the midpoint O is $5.4 \times 10^6 \mathrm{~N/C}$, directed from charge $q_A$ towards charge $q_B$. (b) The force experienced by the test charge is $8.1 \times 10^{-3} \mathrm{~N}$, directed from charge $q_B$ towards charge $q_A$.

The system described is an electric dipole, which consists of two equal and opposite charges separated by a distance.

The total charge of a system is the algebraic sum of all the individual charges. $Q_{total} = q_A + q_B$ $Q_{total} = (2.5 \times 10^{-7} \mathrm{C}) + (-2.5 \times 10^{-7} \mathrm{C})$ $Q_{total} = 0 \mathrm{~C}$

The total charge of the electric dipole is zero.

The electric dipole moment ($\mathbf{p}$) is a vector quantity whose magnitude is the product of the magnitude of either charge ($q$) and the distance of separation between the charges ($2a$). Its direction is from the negative charge to the positive charge.

The torque ($\tau$) experienced by an electric dipole in a uniform electric field is given by the cross product of the dipole moment vector ($\mathbf{p}$) and the electric field vector ($\mathbf{E}$).

Formula: $\mathbf{\tau} = \mathbf{p} \times \mathbf{E}$ The magnitude of the torque is given by: $\tau = pE \sin\theta$ where:

Calculation: Substitute the given values into the formula for the magnitude of the torque: $\tau = (4 \times 10^{-9} \mathrm{~C} \mathrm{~m}) \times (5 \times 10^{4} \mathrm{~N} \mathrm{C}^{-1}) \times \sin(30^{\circ})$ We know that $\sin(30^{\circ}) = 0.5$ or $1/2$. $\tau = (4 \times 5) \times 10^{-9+4} \times 0.5$ $\tau = 20 \times 10^{-5} \times 0.5$ $\tau = 10 \times 10^{-5}$ $\tau = 1 \times 10^{-4} \mathrm{~N} \mathrm{~m}$

Answer: The magnitude of the torque acting on the dipole is $1 \times 10^{-4} \mathrm{~N} \mathrm{~m}$.

This problem uses the principle of quantisation of charge.

Formula: The total charge $q$ is related to the number of electrons $n$ by the formula: $q = ne$ We can find the number of electrons by rearranging the formula: $n = \frac{q}{e}$

Calculation: $n = \frac{-3 \times 10^{-7} \mathrm{C}}{-1.6 \times 10^{-19} \mathrm{C}}$ $n = 1.875 \times 10^{12}$ So, approximately $1.875 \times 10^{12}$ electrons are transferred.

Direction of Transfer: Since the polythene piece acquires a negative charge, it has gained an excess of electrons. The wool, by the law of conservation of charge, must have acquired an equal positive charge, meaning it has lost electrons. Therefore, the electrons are transferred from wool to polythene.

Yes, there is a transfer of mass from wool to polythene. Each electron has a finite, though very small, mass.

Calculation of Mass Transferred: The total mass transferred ($M$) is the number of electrons transferred multiplied by the mass of a single electron. $M = n \times m_e$ $M = (1.875 \times 10^{12}) \times (9.1 \times 10^{-31} \mathrm{~kg})$ $M \approx 1.706 \times 10^{-18} \mathrm{~kg}$

This mass is extremely small and not measurable by ordinary means, but a transfer of mass does indeed occur.

Answer: (a) The estimated number of electrons transferred is $1.875 \times 10^{12}$. The transfer is from wool to polythene. (b) Yes, there is a transfer of mass from wool to polythene, with a total mass of approximately $1.706 \times 10^{-18} \mathrm{~kg}$ being transferred.

We use Coulomb's Law to find the force between the two spheres.

Formula: $F = k \frac{|q_A q_B|}{r^2}$

Calculation: $F = (9 \times 10^9) \frac{(6.5 \times 10^{-7}) \times (6.5 \times 10^{-7})}{(0.5)^2}$ $F = (9 \times 10^9) \frac{42.25 \times 10^{-14}}{0.25}$ $F = (9 \times 10^9) \times (169 \times 10^{-14})$ $F = 1521 \times 10^{-5} = 1.521 \times 10^{-2} \mathrm{~N}$ Since both charges are positive, the force is repulsive.

Let the initial charges be $q_A$ and $q_B$, and the initial distance be $r$. The initial force is $F = k \frac{q_A q_B}{r^2}$.

Calculation of the new force ($F'$): $F' = k \frac{|q'_A q'_B|}{(r')^2}$ $F' = k \frac{|(2q_A)(2q_B)|}{(r/2)^2} = k \frac{4 q_A q_B}{r^2/4} = 16 \left( k \frac{q_A q_B}{r^2} \right)$ $F' = 16 \times F$ $F' = 16 \times (1.521 \times 10^{-2} \mathrm{~N})$ $F' = 24.336 \times 10^{-2} \mathrm{~N} \approx 0.243 \mathrm{~N}$

Alternatively, we can calculate directly: $F' = (9 \times 10^9) \frac{(13.0 \times 10^{-7})^2}{(0.25)^2} = (9 \times 10^9) \frac{169 \times 10^{-14}}{0.0625} = 24336 \times 10^{-5} \mathrm{~N} \approx 0.243 \mathrm{~N}$

Answer: (a) The mutual force of electrostatic repulsion is $1.52 \times 10^{-2} \mathrm{~N}$. (b) The new force of repulsion is $0.243 \mathrm{~N}$.

The figure shows a uniform electrostatic field directed from a positive plate (top) to a negative plate (bottom). The electric field lines point downwards.

The force on a charged particle in an electric field is given by $\mathbf{F} = q\mathbf{E}$.

Let us analyze the tracks:

When a charged particle enters a uniform electric field perpendicularly, it follows a parabolic path. The vertical deflection ($y$) after travelling a horizontal distance ($x$) is given by: $y = \frac{1}{2} a t^2$ where $a$ is the vertical acceleration and $t$ is the time taken to travel the horizontal distance $x$.

The acceleration is caused by the electric force: $a = \frac{F}{m} = \frac{qE}{m}$. The time is $t = \frac{x}{v_x}$, where $v_x$ is the initial horizontal velocity.

Substituting these into the equation for deflection: $y = \frac{1}{2} \left( \frac{qE}{m} \right) \left( \frac{x}{v_x} \right)^2 = \left( \frac{E x^2}{2 v_x^2} \right) \left( \frac{q}{m} \right)$ Assuming all particles enter with the same horizontal velocity ($v_x$) and travel the same horizontal distance ($x$) within the field, the vertical deflection $y$ is directly proportional to the charge-to-mass ratio ($q/m$). $y \propto \frac{q}{m}$

From the figure, for the same horizontal displacement, particle 3 shows the maximum vertical deflection. Therefore, particle 3 has the highest charge to mass ratio.

The electric flux ($\phi$) through a planar surface is defined as the dot product of the electric field vector ($\mathbf{E}$) and the area vector ($\mathbf{S}$).

Formula: $\phi = \mathbf{E} \cdot \mathbf{S} = ES \cos\theta$ where $S$ is the magnitude of the area, and $\theta$ is the angle between the electric field vector and the normal to the area.

If the plane of the square is parallel to the yz plane, its normal vector will be parallel to the x-axis. We can choose the normal to point in the positive x-direction, so its unit vector is $\hat{\mathbf{n}} = \hat{\mathbf{i}}$.

The area vector is $\mathbf{S} = S \hat{\mathbf{n}} = 0.01 \hat{\mathbf{i}} \mathrm{~m}^2$.

The angle $\theta$ between $\mathbf{E}$ (along x-axis) and $\mathbf{S}$ (along x-axis) is $0^{\circ}$.

Calculation: $\phi = ES \cos(0^{\circ})$ $\phi = (3 \times 10^3 \mathrm{~N/C}) \times (0.01 \mathrm{~m}^2) \times 1$ $\phi = 30 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}$ Alternatively, using the dot product: $\phi = \mathbf{E} \cdot \mathbf{S} = (3 \times 10^3 \hat{\mathbf{i}}) \cdot (0.01 \hat{\mathbf{i}}) = 30 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}$

Here, the angle between the normal to the plane and the x-axis is given directly. Since the electric field is along the x-axis, this is the angle $\theta$ between $\mathbf{E}$ and the normal vector.

Calculation: $\phi = ES \cos\theta$ $\phi = (3 \times 10^3 \mathrm{~N/C}) \times (0.01 \mathrm{~m}^2) \times \cos(60^{\circ})$ We know that $\cos(60^{\circ}) = 0.5$. $\phi = 30 \times 0.5$ $\phi = 15 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}$

Answer: (a) The flux through the square is $30 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}$. (b) The flux through the square is $15 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}$.

The problem asks for the net electric flux through a closed surface (a cube) placed in a uniform electric field.

There are two ways to solve this:

Gauss's law states that the net electric flux through any closed surface is equal to the net charge enclosed by the surface divided by the permittivity of free space ($\varepsilon_0$). $\phi_{net} = \frac{q_{enclosed}}{\varepsilon_0}$ A uniform electric field implies that there are no source charges within the region of the field. Therefore, the net charge enclosed by the cube is zero ($q_{enclosed} = 0$).

According to Gauss's Law: $\phi_{net} = \frac{0}{\varepsilon_0} = 0$

Let the cube have six faces. Since the electric field is only in the x-direction ($\hat{\mathbf{i}}$), the flux through the faces that are parallel to the xz and xy planes will be zero. This is because their area vectors are perpendicular to the electric field (along $\pm\hat{\mathbf{j}}$ and $\pm\hat{\mathbf{k}}$ directions).

We only need to consider the two faces perpendicular to the x-axis (the left and right faces).

Net Flux: The net flux is the sum of the fluxes through all six faces. $\phi_{net} = \phi_L + \phi_R + 0 + 0 + 0 + 0$ $\phi_{net} = -120 + 120 = 0$

Both methods show that the net flux is zero. This is a general result: the net electric flux through any closed surface in a uniform electric field is always zero.

Answer: The net flux of the uniform electric field through the cube is zero.

This question can be answered using Gauss's Law.

Gauss's Law: The net electric flux ($\phi$) through a closed surface is directly proportional to the net charge ($q_{net}$) enclosed by that surface. $\phi = \frac{q_{net}}{\varepsilon_0}$ We can rearrange this to find the net charge: $q_{net} = \phi \times \varepsilon_0$

Calculation: $q_{net} = (8.0 \times 10^3) \times (8.854 \times 10^{-12})$ $q_{net} = 70.832 \times 10^{-9} \mathrm{C}$ $q_{net} \approx 7.08 \times 10^{-8} \mathrm{C} \approx 0.07 \mu \mathrm{C}$

So, the net charge inside the box is approximately $0.07 \mu \mathrm{C}$.

If the net outward flux through the surface of the box were zero, we could not conclude that there were no charges inside the box.

Reasoning: According to Gauss's Law, $\phi = q_{net} / \varepsilon_0$. If $\phi = 0$, then it only implies that the net charge inside the box is zero ($q_{net} = 0$).

This condition ($q_{net} = 0$) can arise in two scenarios:

Therefore, a zero net flux only tells us that the total positive charge inside the box is equal to the total negative charge. It does not rule out the presence of charges.

Answer: (a) The net charge inside the box is approximately $7.08 \times 10^{-8} \mathrm{C}$. (b) No, we could not conclude that there were no charges inside the box. We can only conclude that the net charge inside the box is zero. The box could contain equal amounts of positive and negative charge.

This problem can be solved by applying Gauss's Law and using symmetry arguments.

Applying the Hint: We can imagine the square as one face of a cube with an edge length of 10 cm. The charge is located at a distance of 5 cm above the center of this square face. Since 5 cm is half the edge length of the imagined cube (10 cm), the point charge $q$ is located exactly at the center of this cube.

Using Gauss's Law: Gauss's law states that the total electric flux ($\phi_{total}$) through a closed surface is equal to the net charge enclosed ($q_{enclosed}$) divided by the permittivity of free space ($\varepsilon_0$). $\phi_{total} = \frac{q_{enclosed}}{\varepsilon_0}$

For our imaginary cube, the enclosed charge is $q = 10 \times 10^{-6} \mathrm{C}$. So, the total flux through all six faces of the cube is: $\phi_{total} = \frac{10 \times 10^{-6} \mathrm{C}}{8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}} \approx 1.129 \times 10^6 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}$

Using Symmetry: Since the charge is located at the exact center of the cube, the electric field is symmetric with respect to all six faces of the cube. Therefore, the electric flux will be distributed equally among the six faces.

The flux through one face of the cube (which is our original square) will be one-sixth of the total flux. $\phi_{square} = \frac{\phi_{total}}{6}$

Calculation: $\phi_{square} = \frac{1}{6} \times \frac{q}{\varepsilon_0}$ $\phi_{square} = \frac{1}{6} \times \frac{10 \times 10^{-6}}{8.854 \times 10^{-12}}$ $\phi_{square} \approx \frac{1.129 \times 10^6}{6} \approx 1.88 \times 10^5 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}$

Answer: The magnitude of the electric flux through the square is approximately $1.88 \times 10^5 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}$.

This problem is a direct application of Gauss's Law.

Gauss's Law: The net electric flux ($\phi_{net}$) through any closed surface is equal to the net charge enclosed ($q_{enclosed}$) by the surface, divided by the permittivity of free space ($\varepsilon_0$). $\phi_{net} = \frac{q_{enclosed}}{\varepsilon_0}$

An important aspect of Gauss's Law is that the total flux depends only on the amount of enclosed charge, not on the shape or size of the closed (Gaussian) surface.

Calculation: We substitute the given values into Gauss's Law: $\phi_{net} = \frac{2.0 \times 10^{-6} \mathrm{C}}{8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}}$ $\phi_{net} \approx 0.2259 \times 10^6 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}$ $\phi_{net} \approx 2.26 \times 10^5 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}$

Answer: The net electric flux through the surface of the cube is approximately $2.26 \times 10^5 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}$.

According to Gauss's Law, the total electric flux through a closed surface depends only on the net charge enclosed by the surface. It is independent of the size or shape of the Gaussian surface. $\phi = \frac{q_{enclosed}}{\varepsilon_0}$ Since the charge enclosed remains the same, doubling the radius of the spherical Gaussian surface will have no effect on the net electric flux passing through it.

Therefore, the flux would still be $-1.0 \times 10^{3} \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}$.

We can find the value of the point charge using Gauss's Law.

Formula: $\phi = \frac{q}{\varepsilon_0}$ Rearranging for the charge $q$: $q = \phi \times \varepsilon_0$

Calculation: $q = (-1.0 \times 10^3) \times (8.854 \times 10^{-12})$ $q = -8.854 \times 10^{-9} \mathrm{C}$ This can also be written as: $q \approx -8.85 \mathrm{~nC}$

The negative sign of the flux indicates that the enclosed charge is negative, as the electric field lines are directed inward through the surface.

Answer: (a) If the radius of the Gaussian surface were doubled, the flux passing through the surface would remain the same, $-1.0 \times 10^{3} \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}$. (b) The value of the point charge is approximately $-8.85 \times 10^{-9} \mathrm{C}$ (or -8.85 nC).

For a point outside a charged conducting sphere, the electric field is the same as if the entire charge of the sphere were concentrated at its center.

Formula: The magnitude of the electric field ($E$) at a distance ($r$) from the center of a sphere with charge ($q$) is given by: $E = k \frac{|q|}{r^2}$ where $k = 9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}$.

We can rearrange this formula to solve for the charge $q$: $|q| = \frac{E r^2}{k}$

Determining the Sign of the Charge: The electric field points radially outward from a positive charge and radially inward towards a negative charge. Since the field is given to be radially inward, the net charge on the sphere must be negative.

Calculation of the Magnitude of the Charge: $|q| = \frac{(1.5 \times 10^3 \mathrm{~N/C}) \times (0.2 \mathrm{~m})^2}{9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}}$ $|q| = \frac{(1.5 \times 10^3) \times 0.04}{9 \times 10^9}$ $|q| = \frac{0.06 \times 10^3}{9 \times 10^9} = \frac{60}{9 \times 10^9}$ $|q| = 6.66... \times 10^{-9} \mathrm{C}$ $|q| \approx 6.67 \times 10^{-9} \mathrm{C}$