Question

Question 3 Three point charges are arranged as shown. The voltage at point P due to the +Q charge is Vo What is the total voltage at point P? +20 +Q0 7Vc 5% 510

Answer 1

Voltage at point p due to + theta is v_1 = v_0 implies v_1 = k theta/d = v_0 implies theta/d = v_0/k voltage at point p due to - theta charge Is v_2 = k (- theta)/2 d v_2 = k x/2 (- v_0/k) (therefore theta/d = v_0/k) v_2 = -v_0/2 voltage at point p due to + 2 theta charge is v_b = k (2 theta)/d v_3 = 2 k Times v_0/k v_3 = 2 v_0 therefore voltage at point p is v = v_1 + v_2 + v_3 implies v = v_0 - v_0/2 + 2 v_0 implies v = 5/2 v_0 \r\n v_b - v_a = -integral_a^b e dl vector here dl vector = (3 - 00 i^+ (2 - 0)j^= 3 i^+ 2 j^e vector = 5 n/c i^therefore v_a - b_a = -[(5 i^) (3 i^+ 2 j^)] = -5 Times 3 v_b - v_a = -15 volts