5. 4.20 nC and q,--6.40 nC are located at two corners of an equilateral triangle

Question

5. 4.20 nC and q,--6.40 nC are located at two corners of an equilateral triangle Two point charges (q, of edge length 12.0 cm. Point b is located at the third corner, point o is a distance D-600 cm to the left of q1 and a distance of 15.9 cm from q2. a) What is the electric potential at a point b? b) What is the electric potential at a point a? c) What is the electric potential difference between the points a and b? Which point is at a lower electric potential? How much work must be done by an external agent to move a +1.50 nC point charge from point a to point b? d) 9 2 91

Explanation / Answer

a) Electric potential at point b is

               Vb = kq1/L + kq2/L = (9*10^9/0.12)*[4.2 – 6.4]*10^-9 = -165 V

b) Electric potential at point a is

               Va = kq1/D + kq2/d = (9*10^9*4.2*10^-9/0.06) – (9*10^9*6.4*10^-9/0.159) = 267.7 V

c) Vab = Va – Vb = 267.7 + 165 = 432.7 V

d) Work done = W = q*Vab = 1.5*10^-9*432.7 = 649.05 nJ

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