Figure shows a triangular array of three point charges. The electric potential $V$ of these source charges at the midpoint $P$ of the base of the triangle is [image:1_image20.emf] $\left\lbrack \frac{1}{4\pi\epsilon_{0}} = 9 imes 10^{9}Nm^{2}C^{- 2} ight brack$

Question

Figure shows a triangular array of three point charges. The electric potential $V$ of these source charges at the midpoint $P$ of the base of the triangle is [image:1_image20.emf] $\left\lbrack \frac{1}{4\pi\epsilon_{0}} = 9 	imes 10^{9}Nm^{2}C^{- 2} ightbrack$

Dr. Abubacker · Accepted Answer

(b) $V_{P} = \frac{1}{4\pi\varepsilon_{0}}\left\lbrack \frac{q_{1}}{0.2} + \frac{q_{2}}{0.2} + \frac{q_{3}}{0.3} ightbrack$
$= 9 	imes 10^{9}\left\lbrack \frac{1}{0.2} - \frac{2}{0.2} + \frac{3}{0.3} ightbrack 	imes 10^{- 6}$
$= 9 	imes 10^{3}\lbrack 5 - 10 + 10brack = 45 	imes 10^{3} = 45kV$

Q. Figure shows a triangular array of three point charges. The electric potential $V$ of these source charges at the midpoint $P$ of the base of the triangle is

$\left\lbrack \frac{1}{4\pi\epsilon_{0}} = 9 \times 10^{9}Nm^{2}C^{- 2} \right\rbrack$

Explanation

Reviewed by Dr. Abubacker

Want more practice?

Q. Figure shows a triangular array of three point charges. The electric potential $V$ of these source charges at the midpoint $P$ of the base of the triangle is $\left\lbrack \frac{1}{4\pi\epsilon_{0}} = 9 \times 10^{9}Nm^{2}C^{- 2} \right\rbrack$

Explanation

Reviewed by Dr. Abubacker

Want more practice?

Q. Figure shows a triangular array of three point charges. The electric potential $V$ of these source charges at the midpoint $P$ of the base of the triangle is

$\left\lbrack \frac{1}{4\pi\epsilon_{0}} = 9 \times 10^{9}Nm^{2}C^{- 2} \right\rbrack$