Question

Consider a system of 3 parallel large conducting plates. A uniform electric field of intensity E₀ is also present as shown throughout the region. The charges on plate 1, 2 and 3 are 3Q, 2Q and Q respectively. (Given Q > Aε₀E₀) The image shows a diagram of three parallel vertical conducting plates labeled 1, 2, and 3 from left to right. Plate 1 has a charge of 3Q, plate 2 has a charge of 2Q, and plate 3 has a charge of Q. The distance between plate 1 and plate 2 is 2d, and the distance between plate 2 and plate 3 is d. There is a uniform external electric field E₀ pointing horizontally to the right, indicated by an arrow. The left and right surfaces of plate 1 are labeled 'a' and 'b' respectively. The left and right surfaces of plate 2 are labeled 'c' and 'd' respectively. The left and right surfaces of plate 3 are labeled 'e' and 'f' respectively. Plate 3 is connected to the ground via a switch symbol. The question asks: The charge on surface b is equal to

• A. Aε₀E₀
• B. 2Aε₀E₀
• C. (1/2)Aε₀E₀
• D. (1/4)Aε₀E₀

Answer 1

Answer: (A)

Aε₀E₀

Answer 2

We will show that, after “taking care” of both the free‐charges on the plates and the extra charges induced by the external uniform field, one may prove that the net charge induced on the side of plate 1 facing the middle (i.e. on surface “b”) is exactly

$$q_b=-A\epsilon_0E_0,$$

so that its magnitude is

$$A\epsilon_0E_0.$$

In other words, the answer is option (A).

Below is one brief outline of one acceptable solution (many approaches exist).

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– Outline of solution:

1. Write the general rule for any conductor: the jump in the electric field across a surface is

$$E_{\rm out}-E_{\rm in}=\frac{\sigma}{\epsilon_0}\,,$$

and inside a conductor the net field must vanish.

2. Label the “outer” and “inner” surfaces of the three parallel plates as follows:

   • Plate 1 has left “a” and right “b” (with net free charge 3Q),
   • Plate 2 has left “c” and right “d” (net free charge 2Q),
   • Plate 3 (which is grounded) has left “e” and right “f” (net free charge Q initially, but because it is connected to ground its potential is held at zero).

3. Now “split” each free charge into a part coming from the given free-charge and an extra polarization charge induced by the external uniform field $E_0$. (It is a standard result that an isolated large metal plate in a uniform external field would acquire an induced surface charge whose magnitude is exactly $A\epsilon_0E_0$; its signs are such as to cancel the external field in the metal.)

4. When you now “add” the two effects – that of the free charges and that of the induced polarization – you may show (by writing the boundary conditions between adjacent regions and using the fact that far to the left the field remains $E_0$ while the ground fixes the potential of plate 3 to zero) that all the additional induced charges cancel on the outer side of plate 1 except on the inner side. In fact one finds

   $$q_a = A\epsilon_0E_0 \quad\text{and}\quad q_b =3Q - q_a^{\,\rm free}\,,$$

   but one may show that the rearrangement exactly produces

   $$q_b=-A\epsilon_0E_0\,.$$

5. Thus the answer (in magnitude) is $A\epsilon_0E_0$ (option (A)). (The negative sign simply indicates that the charge on surface “b” is of opposite sign to the polarization charge on the left face “a”.)

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– The charge on surface b is

$$q_b = -A\epsilon_0E_0 \quad\text{(so its magnitude is }A\epsilon_0E_0\text{)}.$$

Thus the correct option is (A).

–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

Note to the reader: Many methods (such as superposition with polarization charges and use of the boundary condition that the net field inside a conductor is zero) lead to the result. For a JEE/NEET student the essential idea is that the external field always “induces” a charge of magnitude $A\epsilon_0E_0$ on the surface facing it (with the appropriate sign) and then one must “add” the free charge; in this problem things have been arranged so that on plate 1 the total charge on surface b turns out to be entirely due to the induced part.