Problem 4: 1 Th ree point charges are fixed in space, forming an equilateral tri

Question

Problem 4: 1 Th ree point charges are fixed in space, forming an equilateral triangle as shown below Each side of the triangle is a length d 5.50 m, and the charges all have equal magnitudes, though different sgns: :16 C,6uC 02 4.1: (10 poinls) Find the electric field (magnitude and direction) at the origin 4.2: (10 points) Find the electric potential fie. "voltage") at the onigin 4.3: (10 points) A proton is released at the origin with an initial velocity of 5 m's in the downward (-y) direction. The proton continues to be repelled by the overall charge of the system. Once it has moved infinitely far away from the system shown, how quickly is the proton moving?

Explanation / Answer

4.1) field due to q1 at 'o' [E10] and field due to q2 at 'o' [E20] are equal in magnitude and opposite in direction.so there will not be any field in x-direction.

Field due to q3 at 'o' [E30] is given by

E = E10 + E20 + E30

E = E30 = k.q3/r 2 j = 9*10 9 * 1.6 * 10 -6 / (3*5.5/2)2j

E = 634.71 j N/C

4.2) V = V10 + V20 + V30

V = k { q1/(d/2) + q2/(d/2) + q3/(3 d/2)}

V = 9*109 * 5.5 * { 2*1.6 + 2*1.6 - (2*1.6/3)} * 10-6

V = 225347.72 Mm/C

4.3) eV + 0.5*m*25 = 0.5*m*v 2

1.6*10-19 * 225347.72 + 0.5*25*1.673*10-27 = 0.5*1.673*10-27*v 2

Velocity = 6.565 * 106 m / s

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