Question

Two charges are arranged as shown in Figure Q1.1. Charge q- -48 pC is located on the positive y axis at y-4.0 cm. Charge q,--100 pC is located on the positive x-axis at x-5.0 cm. Grades eople onferences 9,48 pC hat Time 4.0 cm 17 N 5,0㎝ 9:"-100 pc Figure Q1.1 | Question 1 1 pts Calculate the magnitude (in N/C) of E, the electric field at the origin due to q

Answer 1

Ans, The magnitude of the electric field at a distance d due to a point charge q₁ is given by

$E = \frac{k|q_1|}{d^2}$

To find the magnitude of the electric field E₁we put |q₁|= |-89 pC| = 48pC = $48*10^{-12} }$ C and d = 4.0 cm = 0.04 m

So, the magnitude of E₁

$E_1 = \frac{9*10^9*48*10^{-12}}{0.04^2} = 270 \ N/C$

The magnitude of the electric field E₁ = 270 N/C