Question

The plates of a parallel-plate capacitor are given equal positive charges. What will be the potential difference between the plates? What will be the charges on the facing surfaces and on the outer surfaces?

Answer 1

The potential difference between the plates is zero. Charges on the plates’ inner faces are zero. Charges on the plates’ outer faces are +Q.

Explanation:

• In the question, it is understood that the parallel plate capacitor’s two plates have an equal charge. Both these plates have the same charge.
• At point P, due to face I, the electric field is $E_{1}=\frac{Q-q}{2 A \epsilon_{0}}$
• Due to face II, the electric field at point P is $E_{2}=\frac{q}{2 A e}$
• Due to face III, the electric field is E3 = $-\frac{q}{2 A e}$
• Due to face IV, the electric field is E4 = $-\frac{Q+q}{2 A \epsilon_{0}}$

• Inside the conductor, lies the point P. So, the electric field becomes zero. It is represented as, ∴ E1+E2+E3+E4=0

$\Rightarrow \frac{Q-q}{2 A \epsilon_{0}}+\frac{q}{2 A \epsilon_{0}}-\frac{q}{2 A \epsilon_{0}}-\frac{Q+q}{2 A \epsilon_{0}}=0$

So, Q = 0.  

• Between the plates, the potential difference is shown as: $V=\frac{Q}{C}$    
•  Since both have equal charges, both the plates have same potential.

Answer 2

Answer:

The potential difference between the plates is zero. Charges on the plates’ inner faces are zero. Charges on the plates’ outer faces are +Q.

Explanation:

In the question, it is understood that the parallel plate capacitor’s two plates have an equal charge. Both these plates have the same charge.

At point P, due to face I, the electric field is E_{1}=\frac{Q-q}{2 A \epsilon_{0}}E

1

=

2Aϵ

0

Q−q

Due to face II, the electric field at point P is E_{2}=\frac{q}{2 A e}E

2

=

2Ae

q

Due to face III, the electric field is E3 = -\frac{q}{2 A e}−

2Ae

q

Due to face IV, the electric field is E4 = -\frac{Q+q}{2 A \epsilon_{0}}−

2Aϵ

0

Q+q

Inside the conductor, lies the point P. So, the electric field becomes zero. It is represented as, ∴ E1+E2+E3+E4=0

\Rightarrow \frac{Q-q}{2 A \epsilon_{0}}+\frac{q}{2 A \epsilon_{0}}-\frac{q}{2 A \epsilon_{0}}-\frac{Q+q}{2 A \epsilon_{0}}=0⇒

2Aϵ

0

Q−q

+

2Aϵ

0

q

−

2Aϵ

0

q

−

2Aϵ

0

Q+q

=0

So, Q = 0.

Between the plates, the potential difference is shown as: V=\frac{Q}{C}V=

C

Q