Question

A parallel plate capacitor with plates of area 1 m² each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is \(\left( {{\rm{Take,\;}}{{\rm{\varepsilon }}_0} = 8.85 \times {{10}^{ - 12}}\frac{{{{\rm{C}}^2}}}{{{\rm{N}} - {{\rm{m}}^2}}}} \right)\)

1. 9.85 × 10^-10 C
2. 8.85 × 10^-10 C
3. 7.85 × 10^-10 C
4. 6.85 × 10^-10 C

Answer 1

Correct Answer - Option 2 : 8.85 × 10^-10 C

Concept:

The capacitor formula solved for magnitude of charge is:

Q  = C • V

where

Q - the magnitude of charge on each plate of the capacitor.

C - the "capacitance" of a capacitor, a quality invented for and described by q = CV.

V - the magnitude of the change in potential across the capacitor.

Calculation:

If Q = charge on each plate, then

\({\rm{Q}} = {\rm{CV}} = \frac{{{{\rm{\varepsilon }}_0}{\rm{A}}}}{{\rm{d}}}\cdot{\rm{Ed}}\) 

Q = ε₀ AE

Here, A = 1 m², E = 100 N/C

and \({\rm{\;}}{{\rm{\varepsilon }}_0} = 8.85 \times {10^{ - 12}}\frac{{{{\rm{C}}^2}}}{{{\rm{N}} - {{\rm{m}}^2}}}\)

So, by substituting given values, we get

Q = 8.85 × 10^-12 × 1 × 100 = 8.85 × 10^-10 C