A charge q1 = +5.00 nC is placed at the origin of an xy -coordinate system, and a charge q2 = -2.00 nC is placed on the positive x-axis at x = 4.00 cm. (a) If a third charge q3 = +6.00 nC is now placed at the point x = 4.00 cm, y = 3.00 cm, find the x- and y-components of the total force exerted on this charge by the other two. (b) Find the magnitude and direction of this force.

Solution 66P Step 1 of 8: Given charge configuration has charge q =+ 5 1C is placed at origin in xy-coordinate system, another charge q =22nC is at (0,4 cm) on x-axis and a third charge q =+ 6 n3 is placed at point (4 cm, 3 cm) as shown in the figure below. Here we need to calculate the x and y components of total force exerted on charge q by 3 other two and the magnitude and direction of this force. Step 2 of 8: (a) f a third charge q3 = +6.00 nC is now placed at the point x = 4.00 cm, y = 3.00 cm, find the x- and y-components of the total force exerted on this charge by the other two. Figure below represents the given charge configuration along with the vectors from two charges to q 3 Since the charge q a1d q 3 both are positive, the charge 3 repelled away with force F13 . Similarly the charge q2and q 3re opposite in polarity, the charge q3is attracted towards the charge q 3ith the attractive force F23 as shown in the figure above. Step 3 of 8: The position of charge q 1s (0,0) , charge q 2s (0.04 m,0) and charge q i3 at (0.04m, 0.03m) Vector r 23 Using q 20.04 m,0)and q (0304m, 0.03m) r23= (0.04 0.04)i + (0.03 0)j r23= 0.03 j Whereas Unit vector along the r23 is r23 = j Vector r 13 Using q (0,0)and q (0.04m, 0.03m) 1 3 r13= (0.04 0)i + (0.03 0)j r = 0.04 i + 0.03 j 13 Whereas Unit vector along the r13 is r13 = 4 i +3 j 5 5 Step 4 of 8: The force on charge q b3 charge q 2 q2 3 Using coulomb’s law, F 23= k 2r23 (23) Substituting r = 0.03, k=9 × 10 N.m /C , r2 = j , q = 5 × 10 C , q = 2 × 10 C 9 23 23 1 2 in above equation, 9 2 (5×10 C) (2×10 C) F 23 = (9 × 10 N.m /C ) 2 j (0.03) 3 F 23 =0.12× 10 N j