Question

The figure below shows two points in an E-field:
Point  1 is at (X₁,Y₁) = (3,4) in m, and
Point 2 is at (X₂,Y₂) = (12,9) in m.
The Electric Field is constant, with a magnitude of 74 V/m, and is directed parallel to the +X-axis. The potential at point 2 is 434 V. Calculate the potential at point 1.

Calculate the work required for an external force to move a negative charge of Q = -516 microC at constant speed from point 2 to point 1 as defined above in problem 2.

Answer 1

The coordinates of the points are P1 = (3,4) and P2 = (12,9)

The electric field is E = 74 V/m along positive X axis

The potential at P1 is V2 = 434 V

1)

The electric field can be written as the neagative gradient of the potential

Therefore

E = - $\Delta$ V/ $\Delta$ x

$\Delta$ V = -E* $\Delta$ x

$\Delta$ V = -(74)*(12-3)

V2 - V1 = -666

434 -V1 = -666

V1 = 1100V

potential at point 1 is 1100 V

2)

The magnitude of the charge is q = -516 uC

The work done is

W = $\Delta$ V*q

W = (-666 )*(-516*10^-6 )

W = 0.343 J