A parallel plate capacitor has two square plates with equal and opposite charges. The surface charge densities on the plates are +σ and -σ respectively. In the region between the plates the magnitude of the electric field is

Question

A parallel plate capacitor has two square plates with equal and opposite charges. The surface charge densities on the plates are +σ and -σ respectively. In the region between the plates the magnitude of the electric field is (A)  (σ /2 ε 0)  (B)  (σ /ε 0)  (C) zero (D) None of these

Tardigrade Admin · Accepted Answer

The magnitude of the electric field E=(1/2ε0)[σ (-σ )] E=(2σ /2ε 0) E=(σ /ε0)

Q. A parallel plate capacitor has two square plates with equal and opposite charges. The surface charge densities on the plates are $+ σ$ and $- σ$ respectively. In the region between the plates the magnitude of the electric field is

$\frac{σ}{2 ε _{0}}$

$\frac{σ}{ε _{0}}$

zero

None of these

The magnitude of the electric field
$E = \frac{1}{2 ε _{0}} [σ (- σ)]$
$E = \frac{2 σ}{2 ε _{0}}$
$E = \frac{σ}{ε _{0}}$

Q. A parallel plate capacitor has two square plates with equal and opposite charges. The surface charge densities on the plates are +σ and −σ respectively. In the region between the plates the magnitude of the electric field is

2ε0​σ​

ε0​σ​