Question

calculate the value of two equal charges if they repel each other with a force of 0.1 N when situated 50 cm apart in vacuum.what would be the distance between them if they are placed in an insulating medium of dielectric constant 10?

Answer 1

Solution:

Formula to calculate Electrostatic Force between 2 charges:

$\boxed{ \text{Electric Force} = \dfrac{kq_1q_2}{r^2}}$

Here 'k' in vacuum is given as:

$\boxed{ k = \dfrac{1}{4\pi\epsilon_0} = 9 \times 10^9 \:\:\:N.m^2.C^{-2}}$

Since charges are equal, we can say that q₁ and q₂ are equal in magnitude.

⇒ q₁ = q₂ = q

∴ Substituting in the formula, we get:

$\rightarrow 0.1 N = 9 \times 10^9 \times \dfrac{q^2}{(50\times 10^{-2})^2}\\\\\\\rightarrow 0.1 N = \dfrac{9}{2500} \times 10^{9+4} \times q^2\\\\\\\rightarrow 0.1 N = 0.0036 \times 10^{13} \times q^2\\\\\\\rightarrow q^2 = \dfrac{0.1}{0.0036} \times 10^{-13}\\\\\\\rightarrow q^2 = 2.78 \times 10^{-12}\\\\\\\boxed{ \rightarrow q = 1.67 \times 10^{-6} C\:\:\: or\:\:\: 1.67 \mu C}$

Coming to the second part of the question,

It is given that the value of dielectric constant is 10.

So, the new value of K (Coulomb's Constant) is given as:

→ K' = K / 10

∵ Permittivity becomes 10 times.

Hence applying in the formula we get:

$\rightarrow 0.1 N = \dfrac{K}{10} \times (\dfrac{1.67}{r})^2\\\\\\\rightarrow 0.1 N = 9 \times 10^8 \times \dfrac{2.78}{r^2}\\\\\\\rightarrow r^2 = \dfrac{ 9 \times 10^8 \times 2.78}{0.1}\\\\\\\rightarrow r^2 = 250.2 \times 10^8 \:\:m\\\\\\\boxed{ r = 15.82 \times 10^4\;\:\:m}$

This is the required answer.

Answer 2

Answer:

i) Q1 = Q1 =  Q = 1.67 uC

ii) r = 15.8cm  

Explanation:

As, the magnitude of both charges equal so,  Q1 = Q1 =  Q

r = 50 cm

F = 0.1 N

According to Coulomb’s law  

F = K Q^2/r^2     (in Vaccum)

By putting the value in above equation  and solving for Q, we get  

Q = 1.67 uC

Now, for second part the Coulomb’s law will be  

F = K Q^2/E r^2      

Solving for r,

r = 15.8 cm