Question

A charge, q1 = 50-μC is held fixed at the origin and another charge q2 = -20-μC is held fixed on the x axis at a point x = 3.0 m. (a) Calculate the net electric potential at a point A on x-axis at x = 40 cm. (b) Also calculate the net electric potential at another point B on the x-axis at x = 90 cm due the given source charges. (c) If a third charge, q0 = 10-μC charge is released from rest at A, what is its kinetic energy the instant it passes the point B?

Answer 1

Answer:

(a) V = 1.053*10^6 V

(b) V = 4.133*10^5 V

(c) K = 6.397 J

Explanation:

(a) In order to calculate the electric potential at the point A you use the following formula:

$V=k\frac{q}{r}$    

k: Coulomb's constant = 8.98*10^9 NM^2/C^2

q: charge of the particle that generates the electric field

r: distance to the charge

You take into account the two contributions to the electric potential of the two charges:

$V=V_1+V_2=k(\frac{q_1}{r_1}+\frac{q_2}{r_2})$       (1)

q1: 50μC = 50*10^-6 C

q2: -20μC = -20*10^-6 C

r1 = 40cm = 0.4m

r2 = 3.0 m - 0.4 m =  2.6m

You replace the values of all parameters in the equation (1):

$V=(8.98*10^9Nm^2/C^2)(\frac{50*10^{-6}C}{0.4m}+\frac{-20*10^{-6}C}{2.6m})\\\\V=1.053*10^6V$

The net electric potential at A is 1.053*10^6 V

(b) For a point B at x = 90cm = 0.9m you have:

r1 = 0.9m

r2 = 3.0m - 0.9m = 2.1 m

$V=(8.98*10^9Nm^2/C^2)(\frac{50*10^{-6}C}{0.9m}+\frac{-20*10^{-6}}{2.1m})\\\\V=4.133*10^5V$

The net electric potential at the point B is 4.133*10^5 V

(c) In order to calculate the kinetic energy of the charge q0, you use the following formula:

$K=q_oV'$      (2)

q0 = 10μC

V': potential difference between points A and B

You calculate the potential difference V', and then you replace the values of q0 and V in the equation (2):

$V'=1.053*10^6V-4.133*105V=6.39*10^5V\\\\K=(10*10^{-6}C)(6.39*10^5V)=6.397J$

The kinetic energy of the particle when it crosses the point B from point A is 6.397J