Question

Course Sontents HOMEWORK Set 1 ( due Thurs.... Electric field and force due to. Timer Notes The figure shows two point charges. Calculate the magnitude of the electric field at point P. Use the following data: Q1 Q2=-1.40 pC, d1= 1.30 m, d2= 1.80 m. Prine 1.80 pC 02 P 0 2 ubmit Ansee Tries 0/20 Calculate the size of the force on a charge Q-+1.80 μC placed at P due to the two charges from the previous problem. Sulomit Answer Tries 0/20

Answer 1

Part A)

We know tha
E = kq/r^2 for both charges

Since they are both negative, they will provide opposite direction E field at point P, so we will subtract one from another

From Q2,

E = (9 X 10^9)(1.4 X 10^-6)/(1.8)^2

E = 3888.89 N/C to the left

From Q1,

E = (9 X 10^9)(1.8 X 10^-6)/(1.3)^2

E = 9585.8 N/C to the right

The net E field = 9585.8 - 3888.89 = 5696.91NC

Part B)

Apply F = qE

F = (1.8 X 10^-6)(5696.91)

F = 0.01025 N