PART A: Two point charges of +2q and -4q are placed at the corners of the base o

Question

PART A:

Two point charges of +2q and -4q are placed at the corners of the base of an isosceles triangle. The base of the triangle is 7.00 cm and the sides are 14.0 cm, and the q= 11nC. What is the electric potential at the apex of the triangle? (An isosceles triangle has two sides of equal length.)

PART B:

What is the work required to bring a third charge -33.0 nC from infinity and place it at the apex of the triangle? Assume that the potential is zero at infinity.

Only have two tries for each, so please help out if you can! Thank you!

Explanation / Answer

A) Potential due +2q is V1 = k*q1/r1 = (9*10^9*2*11*10^-9)/(0.14) = 1414.3 V

potential due to -4q is V2= k*q2/r2 = (-9*10^9*4*11*10^-9)/0.14 = -2828.57 V

V = V1+V2 = -2828.57 + 1414.3 = -1414.3 V is the electric potential at the apex of the triangle

B) Work done W = q*V = 33*10^-9*1414.3 = 46.67*10^-6 J

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