Question

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Answer 1

Answer:

Hope it's help you......!!

Answer 2

a) 720 × 10⁹  N/c

b) 15.2 × 10¹¹ N/C

Explanation :

Given :

q₁ = +0.2 C

q₂ = +0.4 C

Distance between charges - d = 0.1 m.

To  Find :

a) The mid point between the charge.

b) A point on the line joining q₁ and q₂ such that it is 0.5 m away from q₂ and 0.15 m away from q₁

Solution :

a) The mid point between the charge.

• Electric field due to q₁ -E₁ :

$\implies \sf \dfrac{1}{4\pi \epsilon_0}\quad \dfrac{0.2}{0.5^2}\\ \\ \\\implies \sf 9\times 10^9 \times\dfrac{0.2}{0.05^2} = 720 \times 10^9\;N/C$

• Electric field due to q₂ - E₂ :

$\implies \sf \dfrac{1}{4\pi \epsilon_0}\quad \dfrac{0.2}{0.05^2}=9\times10^9 \times \dfrac{0.4}{0.05^2}\\ \\ \\\implies \sf 1440 \times 10^9\;N/C$

Resultant Electric field at mid-point - E = $\vec{E}_1+\vec{E}_2$

So, the net electric field is acting in opposite direction,

We have E :

$\implies \sf 1440 \times 10^9-720\times10^9 N\\ \\ \\\implies \sf 720 \times 10^9\;N/C$

$\rule{300}{1.5}$

b) A point on the line joining q₁ and q₂ such that it is 0.5 m away from q₂ and 0.15 m away from q₁

Assumption :

The point P be on the line joining the charges such that it is 0.05 m away from q₂ and 0.15 m away from q₁

Electric field due to q₁ :

$\implies \sf 9\times10^9 \times \dfrac{0.2}{0.15^2}\\ \\ \\\implies \sf 80\times10^9\;N/C$

Electric field due to q₂ :

$\implies \sf 9 \times 10^9 \times \dfrac{0.4}{0.05^2}\\ \\ \\\implies \sf 1440\times10^9\;N/C$

Since, Electric field is acting in the same direction  :

$\implies \sf 1520\times 10^9\\ \\ \\ \implies \bf 15.2 \times 10^11\;N/C$