What is the magnitude of the electric field at point p (at the center of square)+2Q-Qp-Q+Q%24

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What is the magnitude of the electric field at point p (at the center of square)
+2Q
-Q
p
-Q
+Q
%24

Expert Solution

Step 1

Given:

Point charges as shown in the figure.

Step 2

Charges at point B and point D is same, so the electric field due to point charge -Q will cancel out and the net electric field will be due to charge at A and C.

The equation for electric filed due to a point charge can be given by,

$E = \frac{k |q|}{r^{2}}$

The horizontal component of electric field due to point charge at A can be given by,

$E_{A H} = \frac{k |q_{A}|}{r^{2}} \cos θ$

Substitute $\frac{s}{\sqrt{2}}$ for r, +2Q for q_A, 9.0 x 10⁹ N•m²/C²for k and 45^o for θ to obtain horizontal component of electric field.

$E_{A H} = \frac{9.0 x 10^{9} {N·m}^{2} / C^{2} |2 Q|}{{(S / \sqrt{2})}^{2}} \cos 45^{o} E_{A H} = \frac{2 \times 9.0 x 10^{9} {N·m}^{2} / C^{2} × |2 Q|}{S^{2}} \times \frac{1}{\sqrt{2}} E_{A H} = \frac{25.45 Q x 10^{9} {N·m}^{2} / C^{2}}{S^{2}}$

Step 3

The vertical component of electric field due to point charge at A can be given by,

$E_{A V} = - \frac{k |q_{A}|}{r^{2}} \sin θ$

Substitute $\frac{s}{\sqrt{2}}$ for r, +2Q for q_A, 9.0 x 10⁹ N•m²/C²for k and 45^o for θ to obtain horizontal component of electric field.

$E_{A V} = - \frac{9.0 x 10^{9} {N·m}^{2} / C^{2} |2 Q|}{{(S / \sqrt{2})}^{2}} \sin 45^{o} E_{A V} = - \frac{2 \times 9.0 x 10^{9} {N·m}^{2} / C^{2} × |2 Q|}{S^{2}} \times \frac{1}{\sqrt{2}} E_{A V} = - \frac{25.45 Q x 10^{9} {N·m}^{2} / C^{2}}{S^{2}}$

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What is the magnitude of the electric field at point p (at the center of square) +2Q -Q p -Q +Q %24