Question

Consider the following figure +4 t4 +24 (a) For the arrangement of charges in the figure, calculate the electric potential (in V) at point P. Use q 9.0 nC and d 5.4 m, and assume that V = 0 V at infinite distance 72.43XV (b) If a charge -2q is brought to point P, what is the electric energy of this charge (in V)? Assume again that the electric potential energy is zero at infinite distance. 1.310-6 x

Answer 1

q = 9 nc d = 5.4 m Here this is a right angle triangle angle ABC so, (AC)^2 = (AB)^2 + (BC)^2 (AC)^2 = d^2 + d^2 AC = squareroot 2 d P is the perpendicular bisector of AC, so, AP = PC = squareroot 2 d/2 = d/squareroot 2 Also, from angle ABP, we can find r r = d cos 45 degree = d/squareroot 2 So, potential at P is given by v_p = dq/d/squareroot 2 + kq/d/squareroot 2 + k (2q)/d/squareroot 2 v_p = squareroot 2 kq/d [1 + 1 + 2] = 4 squareroot 2 kg/d By putting all the values, we get v_p = 4 times 1.414 times 9 times 10^9 times 9 times 10^-9/5.4 = 84.84 v v_p = 84.84 v q_1 = 29 is brought at P electric energy E = qv E = -2 times 9 times 10^-9 times 84.84 E = 1.527 times 10^-6 J