Question 874: Electrostats & Capacitance

Question

Two charged metallic spheres are joined by a very thin metal wire. If the radius of the larger sphere is four times that of the smaller sphere, the electric field near the larger sphere is

Accepted Answer

When two charged metallic spheres are joined by a very thin metal wire, they come to the same electric potential because the wire itself is a conductor and will ensure that the potential difference between the two spheres becomes zero. Let's denote the radii of the smaller and larger spheres as $ R_s $ and $ R_L $, respectively. According to the problem, the radius of the larger sphere is four times that of the smaller sphere:
$$ R_L = 4R_s $$
When the spheres reach electrostatic equilibrium, they share the same electric potential. The electric potential $ V $ on the surface of a sphere is given by:
$$ V = \frac{q}{4\pi \epsilon_0 R} $$
where $ q $ is the charge on the sphere, $ \epsilon_0 $ is the permittivity of free space, and $ R $ is the radius of the sphere.
Let the charges on the smaller and larger spheres be $ q_s $ and $ q_L $. Since the spheres are at the same potential, we have:
$$ \frac{q_s}{4\pi \epsilon_0 R_s} = \frac{q_L}{4\pi \epsilon_0 R_L} $$
Eliminating the constants and solving for the charges, we get:
$$ \frac{q_s}{R_s} = \frac{q_L}{R_L} $$
Since $ R_L = 4R_s $, we can substitute and rearrange to find:
$$ q_L = 4q_s $$
Now, the electric field $ E $ near the surface of a sphere is given by:
$$ E = \frac{q}{4\pi \epsilon_0 R^2} $$
For the smaller sphere, the electric field near its surface is:
$$ E_s = \frac{q_s}{4\pi \epsilon_0 R_s^2} $$
For the larger sphere, substituting $ q_L = 4q_s $ and $ R_L = 4R_s $ into the electric field equation gives:
$$ E_L = \frac{4q_s}{4\pi \epsilon_0 (4R_s)^2} $$
Simplifying, we get:
$$ E_L = \frac{4q_s}{4\pi \epsilon_0 \cdot 16R_s^2} = \frac{q_s}{16\pi \epsilon_0 R_s^2} $$
$$ E_L = \frac{E_s}{4} $$
Therefore, the electric field near the larger sphere is a quarter of that near the smaller sphere.
The correct answer is:
Option B: Quarter of that near the smaller sphere

Answer

twice that near the smaller sphere

Answer

quarter of that near the smaller sphere

Answer

same as that near the smaller sphere

Answer

half of that near smaller sphere

Electrostats & Capacitance

Two charged metallic spheres are joined by a very thin metal wire. If the radius of the larger sphere is four times that of the smaller sphere, the electric field near the larger sphere is

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