Question 842: Electrostats & Capacitance

Question

A conducting sphere of radius $$0.1 \mathrm{~m}$$ has uniform charge density $$1.8 \mu \mathrm{C} / \mathrm{m}^2$$ on its surface. The electric field in free space at radial distance $$0.2 \mathrm{~m}$$ from a point on the surface is ( $$\varepsilon_0=$$ permittivity of free space)

Accepted Answer

To find the electric field at a distance of $$0.2 \mathrm{~m}$$ from the surface of the conducting sphere, we can use Gauss's Law. For a spherical charge distribution, the electric field at an external point is the same as if all the charge were concentrated at the center of the sphere.
First, we calculate the total charge $$Q$$ on the sphere. The charge density $$\sigma$$ is the charge per unit area, so multiplying it by the surface area $$A$$ gives us the total charge $$Q$$.
The surface area of a sphere is given by:
$$A = 4\pi r^2$$
Where $$r$$ is the radius of the sphere. Plugging in the given radius of $$0.1 \mathrm{~m}$$,
$$A = 4\pi (0.1 \mathrm{~m})^2 = 0.04\pi \mathrm{~m}^2$$
The charge density $$\sigma$$ is given as $$1.8 \mu \mathrm{C} / \mathrm{m}^2$$, which is $$1.8 	imes 10^{-6} \mathrm{C} / \mathrm{m}^2$$. Now compute the total charge $$Q$$:
$$Q = \sigma \cdot A = (1.8 	imes 10^{-6} \mathrm{C} / \mathrm{m}^2)(0.04\pi \mathrm{~m}^2)$$
$$Q = (1.8 	imes 10^{-6}) \cdot (0.04\pi) \mathrm{C} = 7.2 	imes 10^{-8} \pi \mathrm{C}$$
Gauss's Law states that the electric field $$E$$ multiplied by the surface area of an imaginary sphere $$A'$$ that encloses the charge is equal to the total charge $$Q$$ enclosed divided by the permittivity $$\varepsilon_0$$:
$$E \cdot A' = \frac{Q}{\varepsilon_0}$$
For a radial distance of $$0.2 \mathrm{~m}$$ from the surface, the actual radial distance from the center of the sphere will be the sum of the sphere's radius (0.1 m) and the distance from the surface (0.2 m), giving us:
$$R = 0.1 \mathrm{~m} + 0.2 \mathrm{~m} = 0.3 \mathrm{~m}$$
The area of the Gaussian surface $$A'$$ is now:
$$A' = 4\pi (0.3 \mathrm{~m})^2 = 0.36\pi \mathrm{~m}^2$$
Plugging the values back into the equation we have:
$$E = \frac{7.2 	imes 10^{-8} \pi \mathrm{~C}}{\varepsilon_0 \cdot 0.36\pi \mathrm{~m}^2}$$
$$E = \frac{7.2 	imes 10^{-8}}{0.36\varepsilon_0}$$
$$E = \frac{7.2}{0.36} 	imes \frac{10^{-8}}{\varepsilon_0}$$
$$E = 20 	imes \frac{10^{-8}}{\varepsilon_0} \mathrm{Vm}^{-1}$$
$$E = 2 	imes \frac{10^{-7}}{\varepsilon_0} \mathrm{Vm}^{-1}$$
Now, the value obtained is that the electric field $$E$$ at a radial distance of $$0.2 \mathrm{~m}$$ from the sphere's surface is equal to:
$$E = \frac{2 	imes 10^{-7}}{\varepsilon_0} \mathrm{Vm}^{-1}$$
This matches Option C.

Answer

$$\frac{6 	imes 10^{-6}}{\varepsilon_0} \mathrm{Vm}^{-1}$$

Answer

$$\frac{6 	imes 10^{-8}}{\varepsilon_0} \mathrm{Vm}^{-1}$$

Answer

$$\frac{2 	imes 10^{-7}}{\varepsilon_0} \mathrm{Vm}^{-1}$$

Answer

$$\frac{1 	imes 10^{-7}}{\varepsilon_0} \mathrm{Vm}^{-1}$$

Electrostats & Capacitance

A conducting sphere of radius $0.1 \mathrm{~m} has uniform charge density 1.8 \mu \mathrm{C} / \mathrm{m}^2 on its surface. The electric field in free space at radial distance 0.2 \mathrm{~m} from a point on the surface is ( \varepsilon_0=$ permittivity of free space)

Choose the correct answer: