Two charged particles with charges q_1 = +31 nC and q_2 = -68 nC and two points

Question

Two charged particles with charges q_1 = +31 nC and q_2 = -68 nC and two points P and S in space are located as shown in the diagram below. Each box in the grid is 0.50 cm by 0.50 cm. (a) What is the value of the electric potential at point P? (b) A third particle with charge -47 nC and mass 1.8 mug is placed at point P and given an initial velocity of 100 directly toward charge It is later observed to pass through point S. What is the speed of this third charge as it passes point S? Will this third particle ever reach a point infinitely far from charges q_1, and q_2? If yes, what is its speed at that point? If no, to what value of the electric potential will the particle reach before turning around?

Explanation / Answer

distance of P from q1, r1 = 6 x 0.50 = 3 cm = 0.03 m

distance of P from q2, r2 = sqrt(6^2 + 8^2) x 0.5 = 5cm = 0.05 m

distance of q1 and q2 from S , r = 4 x 0.5 = 2 cm = 0.02 m

(A) VP = k q1 / r1 + k q2 / r2

= (9 x 10^9) [ (31 x 10^-9 / 0.03) + (-68 x 10^-9 / 0.05)]

= - 2940 Volt

(b) VS = k q1 / r + k q2 / r

= (9 x 10^9) [ (31 x 10^-9 / 0.02) + (-68 x 10^-9 / 0.02)]

= - 16650 Volt

Work done by electric force = change in KE

- (47 x 10^-9) (16650- 2940) = (1.8 x 10^-6) (v^2 - 100^2) / 2

- 6.4437 x 10^-4 = 0.9 x 10^-6 ( v^2 - 100^2)

-715.97 = v^2 - 100^2

v = 96.35 m/s

Initial PE = q V = (-47 x 10^-9) (-2940)

= 1.3818 x 10^-4 J

So yes , it will reach at inifinitely far from charge.

from energy conservation,

1.3818 x 10^-4 + (1.8x 10^-6) (100^2) /2 = 0 + (1.8 x 10^-6)v^2 /2

v = 100.76 m/s

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