Question

Two
equal and similar charges one meter apart
repel eachother with a
force of
one kilogram weight.
Find
the
charges. (g = 10m/s2).​

Answer 1

Answer:

3.33 × 10⁻⁵ C.

Explanation:

Let the charge = q

distance between them, r = 1m

repulsive force between them = force of  one kilogram weight

$\Rightarrow \frac{1}{4 \pi \epsilon_o} \frac{q \times q}{r^2} = mg\\ \\ \Rightarrow \frac{1}{4 \pi \epsilon_o} \frac{q^2}{1^2} = 1 \times 10\\ \\ \Rightarrow 9 \times 10^9\ q^2=10\\ \\ \Rightarrow q^2 = \frac{10}{9 \times 10^9} \\ \\ \Rightarrow q = \sqrt{ \frac{1}{9 \times 10^8} }\\ \\ \Rightarrow q = \frac{1}3 \times 10^{-4}} \\ \\ \Rightarrow \boxed{q = 3.33 \times 10^{-5}\ C}$

Hence the charges are 3.33 × 10⁻⁵ C.

Answer 2

Answer:

Given force =9.8 N

Since q

1

=q

2

as per question.

From coulomb's law

F==

r

2

kq

1

q

2

=

r

2

kq

1

2

q

1

2

= k Fr ²

=1.08×10

−9

q = 1

= 1.08×10

−9 C

Explanation:

Let the charge = q

distance between them, r = 1m

repulsive force between them = force of one kilogram weight

\Rightarrow \frac{1}{4 \pi \epsilon_o} \frac{q \times q}{r^2} = mg\\ \\ \Rightarrow \frac{1}{4 \pi \epsilon_o} \frac{q^2}{1^2} = 1 \times 10\\ \\ \Rightarrow 9 \times 10^9\ q^2=10\\ \\ \Rightarrow q^2 = \frac{10}{9 \times 10^9} \\ \\ \Rightarrow q = \sqrt{ \frac{1}{9 \times 10^8} }\\ \\ \Rightarrow q = \frac{1}3 \times 10^{-4}} \\ \\ \Rightarrow \boxed{q = 3.33 \times 10^{-5}\ C}

Hence the charges are 3.33 × 10⁻⁵ C.