Question

Equipotential surfaces A positive point charge is surrounded by an equipotential surface A, which has a radius of rA. A positive test charge moves from surface A to another equipotential surface B, which has a radius rB. In the process, the electric force does negative work (a) Does the electric force acting on the test charge have the same or opposite direction as the displacement of the test charge? O The electric force has the same direction as the displacement of the test charge. O The electric force has a direction that is opposite to the displacement of the test charge. (b) Is rB greater than or less than Org is greater than rA OrB is less than rA The positive point charge is q-+6.9x10-8 C, and the test charge is go +5.1 x 1011 C. The work done by the electric force as the test charge moves from surface A to surface B is WAB -8.1x10-9 J. The radius of surface A is rA 1.6 m (c) What is the algebraic expression for the potential Ve of surface B? Express your answer in terms of the potential VA of surface A, the work WAB and the test charge go- (d) What is the algebraic expression for the potential V that the point charge q produces at an equipotential surface of radius r? Express your answer in terms of q, r, and the constant k. (e) Find rB, the radius of surface B. Number Unit Select ▼

Answer 1

(a) Does the electric force acting on the test charge have same or opposite direction as the displacement of the test charge?

$\Rightarrow$ The electric force has a direction that is opposite to the displacement of the test charge.

Option (ii) : it will be correct.

(b) Is r_B greater than or less than r_A?

$\Rightarrow$ r_B is less than r_A.

Option (ii) : it will be correct.

(c) The algebraic expression for the potential V_B of surface B which will be given as -

We know that, $\Delta$V_AB = - (W_AB / q₀)

V_B - V_A = (W_AB / q₀)

V_B = V_A + (W_AB / 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 -

V_B - V_A = - W_AB / q₀

(k_e q / r_B) - (k_e q / r_A) = - (W_AB / q₀)

(k_e q / r_B) = (k_e q / r_A) - (W_AB / q₀)

(1 / r_B) = (1 / r_A) - (W_AB / k_e q q₀)

where, r_A = radius of surface A = 1.6 m

W_AB = work done by the electric force from surface A to surface B = - 8.1 x 10^-9 J

k_e = 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 / r_B) = [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 / r_B) = (0.88 m^-1)

r_B = 1 / (0.88 m^-1)

r_B = 1.13 m