Question

Charge q1 = 7. 5 nc is located at the coordinate system origin, while charge q2 = 1. 5 nc is located at (a, 0), where a = 1. 1 m. The point p has coordinates (a, b), where b = 2. 6 m

Answer 1

The electrical field along the horizontal axis is 3.36N/C and the 8.11N/C along the vertical axis

Data;

• q1 = 7.5nC
• q2 = 1.5nC
• a = 1.1m
• b = 2.6m

Electrical Field Force

The electrical field force is given as

$E = \frac{kQ}{r^2}$

The distance apart will be

$r = \sqrt{1.1^2 + 2.6^2} = 2.8m$

Let's find E1 and E2

E1 will be

$E_1 = \frac{-9*10^9 * 7.5*10^-^9}{2.8^2} = -8.61N/C$

E2 will be

$E_2 = \frac{-9*10^9*1.5*10^-^9}{2.8^2} = -1.72N/C$

Let's find the angle between the charges

$tan\theta = \frac{2.6}{1.1} \\\theta = 67.04^0$

The value E12

$E_1_2 = |E|cos\theta = 8.61(cos67.04) = 3.36N/C$

The value of E21 = 0

The value of electrical field along the horizontal axis

$E_x = 3.36 N/c$

The value along the vertical axis

$E_1_y = -|E_1|sin \theta = -7.93 N/C\\E_2_y = E_2 sin 90^0 = -1.72N/C\\E_y = \sqrt{7.93^2 + 1.72^2}= 8.11N/C$

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