Question

kindly help me ....​

Answer 1

$\sf \: {\blue{ \underline{ \underline{ANSWER }}}}$

Given :-

• $\sf \: q_1 \: = \: 5 \times {10}^ {- 8}$
• $\sf \: q_2 = - 3 \times {10}^{ - 8}$
• Distance(d) = 16cm = 0.16m

To Find : -

• Position where the electric potential is 0.

Solution :-

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● 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:-

$\boxed{ \sf \: V = \: \frac{kq_1}{r_1} + \frac{kq_2}{r_2} }$

$\implies \: \sf \: V = \: \frac{kq_1}{x} + \frac{kq_2}{(d - x)}$

When, Electric Potential is zero.

$\implies \: \sf \: 0 = \: \frac{kq_1}{x} + \frac{kq_2}{(d - x)}$

$\implies \: \sf \: \frac{kq_1}{x} = - \frac{kq_2}{(d - x)}$

$\implies \: \sf \: \frac{ \cancel{k}q_1}{x} = - \frac{ \cancel{k}q_2}{(d - x)}$

$\implies \: \sf \: \frac{q_1}{x} = - \frac{q_2}{(d - x)}$

On putting the values,

$\implies \: \sf \: \frac{5 \times {10}^{ - 8} }{x} = \frac{ - ( - 3 \times {10}^{- 8})}{0.16 \: - \: x}$

$\implies \: \sf \: \frac{5 \times \cancel{{10}^{- 8}}}{x} = \frac{ 3 \times \cancel{{10}^{- 8}}}{0.16 \: - \: x}$

On Cross Multiplication, we get:-

$\implies \sf \: \: 3x = \: 0.80 \: - \: 5x$

$\implies \sf \: \: 8x = \: 0.80$

$\implies \: \sf x \: = 0.10 \:$

$\boxed{ \sf \: \therefore \: x =0 .10 \: m \: or \: 10 \: cm}$

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 2

Answer:

Case 1 :- Let two charges 5 × 10⁻⁸C and -3 × 10⁻⁸C are placed on a line of points A answer B and C is the point where electric potential becomes zero.

e.g., A---------------C---------------B

$Then, use </p><p> \: V_C</p><p> =0 \:$

$or, </p><p>V_{CA} +V_{CB} = 0 \:$

$or, </p><p>\frac{q_A}{4\pi\epsilon_0r_{CA}}+\frac{q_B}{4\pi\epsilon_0r_{CB}}=0 \:$

Let CA = x then, CB = (16 - x)

so, 9 × 10⁹ × 5 × 10⁻⁸/x + 9 × 10⁹ × (-3 × 10⁻⁸)/(16 - x) = 0

⇒5/x = 3/(16 - x)

⇒5(16 - x) = 3x

⇒80 - 5x = 3x

⇒80 = 8x

x = 10

Hence, electric potential is zero at the distance of 10cm from charge of 5 × 10⁻⁸C on line joining the two charges between them.

Case 2 :- Let C is not between the two charges. Then,

A-------------B------C

$V_{CA}+V_{CB}=0 \:$

$or, </p><p>\frac{q_A}{4\pi\epsilon_0r_{CA}}+\frac{q_B}{4\pi\epsilon_0r_{CB}}=0 \:$

$or, </p><p>\frac{q_A}{r_{CA}}+\frac{q_B}{r_{CB}}=0 \:$

or, 5 × 10⁻⁸/(16 + x) + (-3 × 10⁻⁸)/x = 0

or, 5/(16 + x) = 3/x

or, 5x = 48 + 3x

or, 2x = 48 ⇒x = 24cm

So, electric potential is also equal to zero at distance of 24cm from charge of - 3 × 10⁻⁸C and at a distance of (24 + 16) cm from charge of 5 × 10⁻⁸C,on the side of charge of - 3 × 10⁻⁸C

hope this helps you