two charges 5×10^(-8) C and -3×10^(-8) are located 16cm apart. at what points on the line joining the two charges is the electric potential zero ? take the potential at infinity to be zero.
Home
/
two charges 5×10^(-8) C and -3×10^(-8) are located 16cm apart. at what points on the line joining the two charges
Verified answer
[tex] \sf \: {\blue{ \underline{ \underline{Answer ~:}}}}[/tex]
Given :
[tex]\sf \: q_1 \: = \: 5 \times {10}^ {- 8}[/tex]
[tex]\sf \: q_2 = - 3 \times {10}^{ - 8}[/tex]
Distance(d) = 16cm = 0.16m
To Find :
Position where the electric potential is 0.
Solution :
[tex] \setlength{\unitlength}{1cm}\thicklines\begin{picture}(10,6)\put(1,1){\vector(-1,0){3}} \put(1,1){\vector(1,0){3}}\put(-1.5,1){\circle*{0.1}}\put(1.5,1){\circle*{0.1}}\put(3.5,1){\circle*{0.1}}\put(-1.5,2){\line(0,1){0.3}}\put(-1.5,2.15){\line(1,0){3}}\put(1.5,2){\line(0,1){0.3}}\put(-1.5,-0.5){\line(0,1){0.3}}\put(-1.5,-0.35){\line(1,0){5}}\put(3.5,-0.5){\line(0,1){0.3}}\put(-1.6,0.5){\sf A}\put(1.35,0.5){\sf B}\put(3.4,0.5){\sf C}\put(-0.2,1.7){\sf x\;cm}\put(0.2,0){\sf d = 16\;cm}\put(-1.6,1.4){$\sf q_1$}\put(3.4,1.4){$\sf q_2$}\end{picture} [/tex]
● 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,
[tex] \boxed{ \sf \: V = \: \frac{kq_1}{r_1} + \frac{kq_2}{r_2} }[/tex]
[tex] \implies \: \sf \: V = \: \frac{kq_1}{x} + \frac{kq_2}{(d - x)}[/tex]
When, Electric Potential is zero.
[tex] \implies \: \sf \: 0 = \: \frac{kq_1}{x} + \frac{kq_2}{(d - x)}[/tex]
[tex] \implies \: \sf \: \frac{kq_1}{x} = - \frac{kq_2}{(d - x)}[/tex]
[tex] \implies \: \sf \: \frac{ \cancel{k}q_1}{x} = - \frac{ \cancel{k}q_2}{(d - x)}[/tex]
[tex] \implies \: \sf \: \frac{q_1}{x} = - \frac{q_2}{(d - x)}[/tex]
On putting the values,
[tex] \implies \: \sf \: \frac{5 \times {10}^{ - 8} }{x} = \frac{ - ( - 3 \times {10}^{- 8})}{0.16 \: - \: x}[/tex]
[tex]\implies \: \sf \: \frac{5 \times \cancel{{10}^{- 8}}}{x} = \frac{ 3 \times \cancel{{10}^{- 8}}}{0.16 \: - \: x}[/tex]
On Cross Multiplication, we get
[tex] \implies \sf \: \: 3x = \: 0.80 \: - \: 5x[/tex]
[tex]\implies \sf \: \: 8x = \: 0.80 [/tex]
[tex] \implies \: \sf x \: = 0.10 \: [/tex]
[tex] \boxed{ \sf \: \therefore \: x =0 .10 \: m \: or \: 10 \: cm}[/tex]
Thus, the Electric Potential is zero at 10cm at right side of charge q(1) or (16 - 10) = 6cm at left side of charge q(2).
Answer:
[tex]\sf{q_{1} = 5 \times 10^{-8}}[/tex]
[tex]\sf{q_{2} = -3 \times 10^{-8}} \\ \\ [/tex]
[tex]\implies\sf{ \frac{5 \times 10^{-8}}{x} = \frac{ -(-3 \times 10^{-8}}{0.16 -x} }[/tex]
[tex]\longrightarrow\sf{ \frac{5 \times \cancel{10^{-8}}}{x} = \frac{ -(-3 \times \cancel{10^{-8}}}{0.16 -x} }[/tex]
[tex]\implies\sf{ 3x = 0.80 - 5x}[/tex]
[tex]\implies\sf{ 3x + 5x = 0.80}[/tex]
[tex]\implies\sf{ 8x = 0.80}[/tex]
[tex]\implies\bf\green{\fbox{ x = 0.10}}[/tex]
Hope it will be helpful ✌️