Use the worked example above to help you solve this problem. A 5.30-mu C point c

Question

Use the worked example above to help you solve this problem. A 5.30-mu C point charge is at the origin, and a point charge q_2 = -1.50 mu C is on the x-axis at (3.00, 0)m, as shown in the figure. (a) If the electric potential is taken to be zero at infinity, find the electric potential due to these charges at point P with coordinates (0, 4.00)m. (b) How much work is required to bring a third point charge of 4.20 mu C from infinity to P? Suppose a charge of -1.50 mu C is at the origin and a charge of 2.90 mu C is at the point (0, 3.00)m. (a) Find the electric potential at (4.00, 0)m, assuming the electric potential is zero at infinity. (b) Find the work necessary to bring a 4.20 mu C charge from infinity to the point (4.00, 0)m.

Explanation / Answer

PRACTICE IT:

(a) Electric potential = k q / r

V = [ k q1 / 4] + [ k q2 / srqt(3^2 + 4^2)]

= (9 x 10^9) [ (5.30 x 10^-6 / 4) + ( - 1.50 x 10^-6 / 5)]

= 9225 Volt

(b) work done = q deltaV

= (4.20 x 10^-6) (9225)

= 0.0387 J

EXERCISE:

(a) Electric potential = k q / r

V = [ k q1 / 4] + [ k q2 / srqt(3^2 + 4^2)]

= (9 x 10^9) [ (-1.50 x 10^-6 / 4) + ( 2.90 x 10^-6 / 5)]

= - 1845 Volt

(b) work done = q deltaV

= (4.20 x 10^-6) ( - 1845)

= -7.75 x 10^-3 J

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