Question

Two oppositely charged parallel metal plates,

Answer 1

Answer: 2.25 × 10⁴ N/C

Explanation:

The magnitude of the electric field strength is given by:

$E = \frac{F}{q}$

where, F is the electrostatic force and q is the charge.

It is given that the magnitude of the force acting on an electron kept between the plates is: F = 3.60 ×10⁻¹⁵ N

Charge of the electron, q = 1.6 × 10⁻¹⁹ C

Thus, E = (3.60 ×10⁻¹⁵ N)÷(1.6 × 10⁻¹⁹ C) = 2.25 × 10⁴ N/C.

Thus, the electric field strength between the plates is 2.25 × 10⁴ N/C.

Answer 2

Electric field strength = Force/Charge

E = (3.6*10^-15)/(1.6*10^-19) = 2.25*10^4 N/c

1.6*10^-19 being the elementary charge that is a fundamental constant, and is the charge on an electron.