Question

Two charges 8 μC and –6 μC are held at certain distance apart in air, attract each other with a force of 12 N. A charge 4 μC is added to each. The electrostatic force between new charges held at same separation in a medium of dielectric constant 2 would be
0 N

12 N

3 N

2 N

Answer 1

Answer:

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Explanation:

Two charges 8 μC and –6 μC are held at certain distance apart in air, attract each other with a force of 12 N. A charge 4 μC is added to each. The electrostatic force between new charges held at same separation in a medium of dielectric constant 2 would be

Answer 2

Answer:

The electrostatic force between new charges held at the same separation in a medium of dielectric constant 2 would be 3 N.

Explanation:

We know that the force between the two charges q1 and q2 separated by distance r is given by:

$F = \frac{1}{4\pi \epsilon_0 \epsilon_r} \frac{q1q2}{r^2}$...(1)

In situation 1:

q1 = 8 μ C, q2 = -6 μC, F = -12 N(as it is attractive), and $\epsilon_r = 1$ (for vacuum)

So,

$-12 = \frac{k(8 \mu C)(-6\mu C)}{r^2 (1)}$  [Here k = 1/(4πε₀)]

⇒ r = √k × 2 × 10⁻⁶C²

Now for situation 2:

We substitute q1 = 12 μC, q2 = -2 μC, r = √k × 2 × 10⁻⁶C², $\epsilon_r=2$(given) in equation (1), we get:

$F = \frac{k(12 \mu C)(-2 \mu C)}{2 \times r^2}$

⇒ F = $\frac{k(12 \mu C)(-2\mu C)}{2\times (\sqrt{k} \times 2 \times \mu C)^2}$

⇒ F = 3 N.

Therefore, the electrostatic force between new charges held at the same separation in a medium of dielectric constant 2 would be 3 N.

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