Suppose a charge -Q1 is given to the positive plate and a charge -Q2 to the negative plate of a capacitor- What is the -charge on the capacitor-

Given: nbsp; Charge on the positive plane = +Q1 nbsp; Charge on the negative plate = Q2 To calculate: Charge on the capacitor Let ABCD be the Gaussian ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Gauss's Law

Suppose a cha...

Question

Suppose a charge +Q₁ is given to the positive plate and a charge −Q₂ to the negative plate of a capacitor. What is the "charge on the capacitor"?

Open in App

Solution

Given:
Charge on the positive plane = +Q₁
Charge on the negative plate = $-$ Q₂

To calculate: Charge on the capacitor

Let ABCD be the Gaussian surface such that faces AD and BC lie inside plates X and Y, respectively.
Let q be the charge appearing on surface II. Then, the distribution of the charges on faces I, III and IV will be in accordance with the figure.

Let the area of the plates be A and the permittivity of the free space be $\in_{0}$ .

Now, to determine q in terms of Q₁ and Q₂, we need to apply Gauss's law to calculate the electric field due to all four faces of the capacitor at point P. Also, we know that the electric field inside a capacitor is zero.

Electric field due to face I at point P, E₁ = $\frac{Q_{1} - q}{2 \in_{0} A}$
Electric field due to face II at point P, E₂ = $\frac{+ q}{2 \in_{0} A}$
Electric field due to face III at point P, E₃ = $\frac{- q}{2 \in_{0} A}$
Electric field due to face IV at point P, E₄ = $- (\frac{- Q_{2} + q}{2 \in_{0} A})$ (Negative sign is used as point P lies on the LHS of face IV.)
Since point P lies inside the conductor,
E₁ + E₂ + E₃ + E₄ = 0

∴ Q₁ $-$ q + q $-$ q $-$ ( $-$ Q₂ + q ) = 0
$\Rightarrow$ q = $\frac{Q_{1} + Q_{2}}{2}$
Thus, the charge on the capacitor is $\frac{Q_{1} + Q_{2}}{2}$ , which is the charge on faces II and III.

Suggest Corrections

Similar questions

Q. Charges

Q_{1}

and

Q_{2}

are given to two plates of a parallel plate capacitor. The capacity of the capacitor is

C

. When the switch is closed, mark the correct statement(s): (Assume both

Q_{1}

and

Q_{2}

to be positive)

Each plate of a parallel plate air capacitor has area

S = 5 \times 10^{- 3} m^{2}

and the distance between the plates is

d = 8.80 m m

. Plate

A

has positive charge

q_{1} = + 10^{- 10} C

, and plate

B

has charge

q_{2} = + 2 \times 10^{- 10} C

. A battery of emf

E = 10 V

has its positive terminal connected to plate

A

and the negative terminal to plate

B

. Charge supplied by the battery is :

(Given

ε_{0} = 8.85 \times 10^{- 12} N m^{2} C^{- 2}

)

Q. A parallel plate capacitor has a plate area

A

, separation

d

and charge

Q

. A positive charge

q

of mass

m

is released from a point very near to the positive plate. The time taken by the charge to reach to the negative plate of capacitor is
(Neglect gravity)

Q. Why the charge on a capacitor always equal to charge on it's positive plate and not on the negative one?

Q. Two parallel plate capacitors A and B having capacitances of

1 μ F

and

5 μ F

are charged separately to the same potential of

100

V. Now the positive plate of A is connected to the negative plate of B and the negative plate of A to the positive plate of B. The final charge on each capacitor is?

Join BYJU'S Learning Program

Suppose a charge +Q1 is given to the positive plate and a charge −Q2 to the negative plate of a capacitor. What is the "charge on the capacitor"?

Suppose a charge +Q₁ is given to the positive plate and a charge −Q₂ to the negative plate of a capacitor. What is the "charge on the capacitor"?