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Two spherical conductors A and B of radi...

Two spherical conductors A and B of radii 1mm and 2mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is

A

`4 :1`

B

`1:2`

C

`2:1`

D

`1:4`

Text Solution

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The correct Answer is:
To solve the problem step by step, we will analyze the situation involving two spherical conductors A and B, their radii, and the electric fields at their surfaces. ### Step 1: Understand the Problem We have two spherical conductors: - Sphere A with radius \( r_1 = 1 \, \text{mm} = 0.001 \, \text{m} \) - Sphere B with radius \( r_2 = 2 \, \text{mm} = 0.002 \, \text{m} \) They are separated by a distance of \( d = 5 \, \text{cm} = 0.05 \, \text{m} \). When connected by a conducting wire, they will reach an equilibrium state where their potentials are equal. ### Step 2: Set Up the Equations When the spheres are connected by a wire, the potentials \( V_A \) and \( V_B \) will be equal: \[ V_A = V_B \] The potential \( V \) of a charged sphere is given by: \[ V = k \frac{Q}{r} \] where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the radius of the sphere. Thus, we can write: \[ k \frac{Q_1}{r_1} = k \frac{Q_2}{r_2} \] ### Step 3: Simplify the Equation Since \( k \) is a constant, it cancels out: \[ \frac{Q_1}{r_1} = \frac{Q_2}{r_2} \] This can be rearranged to give us the ratio of the charges: \[ \frac{Q_1}{Q_2} = \frac{r_1}{r_2} \] ### Step 4: Calculate the Electric Fields The electric field \( E \) at the surface of a charged sphere is given by: \[ E = k \frac{Q}{r^2} \] For spheres A and B, we have: \[ E_A = k \frac{Q_1}{r_1^2} \] \[ E_B = k \frac{Q_2}{r_2^2} \] ### Step 5: Find the Ratio of Electric Fields We want to find the ratio \( \frac{E_A}{E_B} \): \[ \frac{E_A}{E_B} = \frac{k \frac{Q_1}{r_1^2}}{k \frac{Q_2}{r_2^2}} = \frac{Q_1}{Q_2} \cdot \frac{r_2^2}{r_1^2} \] Substituting the ratio of charges from Step 3: \[ \frac{E_A}{E_B} = \frac{r_1}{r_2} \cdot \frac{r_2^2}{r_1^2} \] ### Step 6: Simplify the Expression This simplifies to: \[ \frac{E_A}{E_B} = \frac{r_2}{r_1} \] ### Step 7: Substitute the Values Now substituting the values of \( r_1 \) and \( r_2 \): \[ \frac{E_A}{E_B} = \frac{2 \, \text{mm}}{1 \, \text{mm}} = 2 \] ### Final Answer Thus, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is: \[ \frac{E_A}{E_B} = 2:1 \]
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