(a) Does the electric force acting on the test charge have same or opposite direction as the displacement of the test charge?
The electric force has a direction that is opposite to the displacement of the test charge.
Option (ii) : it will be correct.
(b) Is rB greater than or less than rA?
rB is less than rA.
Option (ii) : it will be correct.
(c) The algebraic expression for the potential VB of surface B which will be given as -
We know that, VAB = - (WAB / q0)
VB - VA = (WAB / q0)
VB = VA + (WAB / q0)
(d) The algebraic expression for the potential V which will be given as -
V = k q / r
where, k = proportionality constant
q = point charge
r = radius of equipotential surface
(e) The radius of surface B which will be given by -
VB - VA = - WAB / q0
(ke q / rB) - (ke q / rA) = - (WAB / q0)
(ke q / rB) = (ke q / rA) - (WAB / q0)
(1 / rB) = (1 / rA) - (WAB / ke q q0)
where, rA = radius of surface A = 1.6 m
WAB = work done by the electric force from surface A to surface B = - 8.1 x 10-9 J
ke = proportionality constant = 9 x 109 Nm2/C2
q = positive point charge = 6.9 x 10-8 C
q0 = test charge = 5.1 x 10-11 C
then, we get
(1 / rB) = [1 / (1.6 m)] + {(8.1 x 10-9 J) / [(9 x 109 Nm2/C2) (6.9 x 10-8 C) (5.1 x 10-11 C)]}
(1 / rB) = (0.88 m-1)
rB = 1 / (0.88 m-1)
rB = 1.13 m
Equipotential surfaces A positive point charge is surrounded by an equipotential surface A, which has a...
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