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

(a) V = 1.053*10^6 V

(b) V = 4.133*10^5 V

(c) K = 6.397 J

Step-by-step explanation:

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

`V=k(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((q_1)/(r_1)+(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)((50*10^(-6)C)/(0.4m)+(-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)((50*10^(-6)C)/(0.9m)+(-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