Question

Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by a distance of 40.0 mm, and the potential difference between them is 370 V
A. What is the magnitude of the electric field (assumed to be uniform) in the region between the plates?
B. What is the magnitude of the force this field exerts on a particle with a charge of 2.40 nC ?
C. Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower.
D. Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.

Answer 1

Answer:

Explanation:

A )

electric field E = V / d where V is potential difference between plates separated by distance d .

putting the given values

E = 370 / .040  V / m

= 9250 V / m

B )

Force on charged particle of charge q in electric field E

F = q E

F = 2.4 x 10⁻⁹ x 9250

= 22200 x 10⁻⁹

= 222 x 10⁻⁷ N .

C ) since field is uniform , force will be constant

work done by electric field putting up this force

= force x displacement

= 222 x 10⁻⁷  x 40 x 10⁻³

= 888 x 10⁻⁹ J

D )

change in potential energy

= q ( V₁ - V₂ )

= 2.40 X 10⁻⁹ x 370

= 888 x 10⁻⁹ J .

Answer 2

(a) The magnitude of electric field in the region between the plates is 9,250 V/m.

(b) The magnitude of the force the field exerts on a particle with the given charge is 2.22 x 10⁻⁵ N.

(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is $8.88 \times 10^{-7} \ J$.

(d) the change of the potential energy is $8.88 \times 10^{-7} \ J$.

The given parameters;

• distance between the two metal plates, d = 40 mm
• potential difference between the plates, V = 370 V

(a) The magnitude of electric field in the region between the plates is calculated as;

$E = \frac{V}{d} \\\\E = \frac{370 }{40 \times 10^{-3} } \\\\E = 9,250 \ V/m$

(b) The magnitude of the force the field exerts on a particle with the given charge is calculated as follows;

F = Eq

F = 9,250 x 2.4 x 10⁻⁹

F = 2.22 x 10⁻⁵ N

(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is calculated as follows;

$W = Fd\\\\W = 2.22 \times 10^{-5} \times 40\times 10^{-3} \\\\W =8.88 \times 10^{-7} \ J$

(d) the change of the potential energy is calculated as;

$\Delta U = q \Delta V\\\\\Delta U = q(V_1 - V_2)\\\\\\Delta U = 2.4 \times 10^{-9}(370)\\\\\Delta U = 8.88 \times 10^{-7} \ J$

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