Question

Two point charges 2×10^-7C and other 1×10^-7 are 1.0cm apart .what is the magnitude of the field produced by the either charge at the side of the other.

Answer 1

E = 9×10^9 × 2×10^-7/(10^-2)^2

E = 126×10^6 =1.26×10^8 V/M

Answer 2

Explanation:

It is given that,

Charge 1, q₁ = 2 × 10⁻⁷ C

Charge 2, q₂ = 1 × 10⁻⁷ C

Distance between charges, d = 1 cm = 0.01 m

The magnitude of the field produced by the either charge at the side of the other is given by :

The electric field is given by :

$E=\dfrac{F}{q_1}$

$E=\dfrac{9\times 10^9\times 2\times 10^{-7}\times 10^{-7}}{(0.01)^2\times 2\times 10^{-7}}$

E = 9000000 N/C

$E=9\times 10^6\ N/C$

Hence, this is the required solution.