Answer:
a. Work done by the electric force = -2.85 * ×10⁻⁵ J
b. The potential of the starting point with respect to the end point = -3.03 * 10³ V
c. The magnitude of the electric field is 33.7kV/m
Step-by-step explanation:
Given.
Charge = Q = 9.40 nC
Distance = d = 9.00 cm = 0.09m
Amount of work = 7.10×10⁻⁵ J
Kinetic energy = K = 4.25×10⁻⁵ J
a. What work was done by the electric force?
This is calculated by; change in Kinetic Energy i.e. ∆KE
∆KE = ∆K2 - ∆Kæ
Where K2 = 4.25×10⁻⁵ J
The body is released at rest, so the initial velocity is 0.
So, K1 = 0
Also, total work done = W1 + W2
Where W2 = 7.10×10⁻⁵J
So, W1 + W2 = W = K2
W1 + 7.10×10⁻⁵ = 4.25×10⁻⁵
W1 = 4.25×10⁻⁵ - 7.10×10⁻⁵
W = -2.85 * ×10⁻⁵ J
Work done by the electric force = -2.85 * ×10⁻⁵ J
b. What is the potential of the starting point with respect to the end point?
The change in potential energy is given as
W = ∆U
W = Q|V2 - V1| where V1 = 0 because the body starts from rest
So, W = QV2
Make V the Subject of the formula
V2 = W/Q
V2 = -2.85 * ×10⁻⁵ J / 9.40 nC
V2 = -2.85 * ×10⁻⁵ J / 9.40 * 10^-9C
V2 = −3031.9148936170212765957V
V2 = -3.03 * 10³ V
The potential of the starting point with respect to the end point = -3.03 * 10³ V
c. What is the magnitude of the electric field?
The magnitude of the electric field is calculated as follows;
W = -Fd = -QEd
And E = V/d
E = -3.03 * 10³ V / 0.09 m
E = −33687.943262411347517730 V/m
E = -33.7kV/m
The magnitude of the electric field is 33.7kV/m