Question

Two charges 3 × 10–8 C and –2 × 10–8 C are located 15 cm apart. At what point on the line

joining the two charges is the electric potential zero? Take the potential at infinity to be zero.​

Answer 1

Explanation:

Given :-

Object distance (u)= -50cm

Radius (R)= -60cm [•°• Concave mirror]

focal length (f)= R/2= -60/2= (-30)cm

Size of object = 15cm

To Find :-

We have to find the size of image

Solution :-

Using mirror formula

\underline{\boxed{\it\ \dfrac{1}{f}= \dfrac{1}{v}+\dfrac{1}{u}}}

f

1

=

v

1

+

u

1

Now , by putting these values in the formula

\begin{gathered}\longrightarrow\sf\ \dfrac{1}{-30}= \dfrac{1}{-50}+\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ \dfrac{-1}{30}+\dfrac{1}{50}=\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ \dfrac{-5+3}{150}=\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ \dfrac{-2}{150}=\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ v= \cancel{\dfrac{-150}{2}}\\ \\ \\ \longrightarrow\purple{\sf v= (-75)cm}\end{gathered}

⟶

−30

1

=

−50

1

+

v

1

⟶

30

−1

+

50

1

=

v

1

⟶

150

−5+3

=

v

1

⟶

150

−2

=

v

1

⟶ v=

2

−150

⟶v=(−75)cm

Now we have to find the size of image

By using magnification formula

\underline{\boxed{\sf\ m= \dfrac{-v}{u}= \dfrac{h_i}{h_o}}}

m=

u

−v

=

h

o

h

i

\begin{gathered}\longrightarrow\sf\ \dfrac{h_i}{15}=\dfrac{\not{-}(-75)}{\not{-}50}\\ \\ \\ \longrightarrow\sf\ \dfrac{h_i}{15}=\dfrac{-(75)}{50}\\ \\ \\ \longrightarrow\sf\ \ h_i= \dfrac{-(\cancel{75}\times 15)}{\cancel{50}}\\ \\ \\ \longrightarrow\sf\ \ h_i= \dfrac{-3\times 15}{2}\\ \\ \\ \longrightarrow\sf\ \ h_i= \cancel{\dfrac{-45}{2}}\\ \\ \\ \longrightarrow\sf\ \ h_i= (-22.5)\end{gathered}

⟶

15

h

i

=



−50



−(−75)

⟶

15

h

i

=

50

−(75)

⟶ h

i

=

50

−(

75

×15)

⟶ h

i

=

2

−3×15

⟶ h

i

=

2

−45

⟶ h

i

=(−22.5)

\underline{\bigstar{\textsf{\textbf{\ Size\ of\ image = (-22.5)cm}}}}

★ Size of image = (-22.5)cm

Answer 2

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

• $q1 = 3 × {10}^{-8}$

• $q2 = -2 x {10}^{-8}$

• 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).

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