Physics, asked by jagannathhansda29, 3 months ago

two charges 3×two charges 3×10-8c and -2×10-8 c are located 15 cm .at what point on the line joining the two charges is the electric potential zero? take the potential at infinity to be zero​

Answers

Answered by BrainlyBAKA
0

\huge\green{\underline{\underline{Given :}}}

\\

  • q1 = 3 × {10}^{-8}

  • q2 = -2 x {10}^{-8}

  • Distance(d) = 15 cm = 0.15m

\\

\huge\green{\underline{\underline{To\:Find :}}}

\\

  • Position where the electric potential is 0.

\\

\huge\green{\underline{\underline{Solution :}}}

\\

\huge{•} Let C be the point from 'x cm' to the right of charge q(1) where electric potential is zero. (as shown in the fig.)

\\

According to Question,

\\

\pink{\boxed{{V \:=\: }\frac{kq1}{r1}+\frac{kq2}{r2}}}

\\

 =>\large{ V \:=\: \frac{kq1}{x}+\frac{kq2}{d-x}}

When, Electric Potential is zero,

 => \large{0 \:=\: \frac{kq1}{x}+\frac{kq2}{d-x}}

 \large{=> \frac{kq1}{x}  \:=\: - \frac{kq2}{d-x}}

 \large{=> \frac{q1}{x}  \:=\: - \frac{q2}{d-x}}

On putting the values,

 \large{=> \frac{3 × {10}^{-8}}{x}  \:=\: - \frac{-2 x {10}^{-8}}{0.15-x}}

 \large{=> \frac{3}{x}  \:=\: \frac{2}{0.15-x}}

On Cross Multiplication we get,

=> 2x\: =\: 0.50\: - \:3x

=> 5x\: =\: 0.50

=> x \:= \:0.10

\\

\large\pink{\fbox{=>\:x\: =\: 0.10\: m\: or\: 10\: cm}}

\\

Thus, the Electric Potential is zero at 10 cm at right side of charge q(1) or (15-10) 5 cm at left side of charge q(2).

\\\\\\

HOPE IT HELPS

PLEASE MARK ME BRAINLIEST ☺️

Similar questions