Answered

Two oppositely charged parallel metal plates,

1.00 centimeter apart, exert a force with a

magnitude of 3.60 × 10^–15 newton on an electron

placed between the plates. Calculate the

magnitude of the electric field strength between

the plates. [Show all work, including the

equation and substitution with units.]

Verified Answer :

TaylorBayley

Electric field strength = Force/Charge

E = (3.610^-15)/(1.610^-19) = 2.2510^4 N/c

1.610^-19 being the elementary charge that is a fundamental constant, and is the charge on an electron.

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.

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