Question

a charge q when placed in an electric field of intensity 50n/c experience of force 5*10000n in air.Find the magnitude of charge (g=10m/s)

Answer 1

Answer:

0.45 × 10^-4

Explanation:

Magnitude of charge E = F/q

F = 2.25N

Force q=5*10^-4

E = 0.45 × 10^-4

Answer 2

CHARGE IS 1000 COULOMB.

Given:

• Electrostatic field intensity is 50 N/C
• Force experienced is 5 × 10000 N

To find:

Magnitude of charge ?

Calculation:

Let's assume that the field intensity vector is uniform across the whole field such that the force experienced by any charge will be constant anywhere in the field.

The general expression between field intensity and force is given as:

$F = E \times q$

• F is force, E is field intensity magnitude, q is charge.

$\implies \: q = \dfrac{F}{E}$

$\implies \: q = \dfrac{5 \times 10000}{50}$

$\implies \: q = 1000 \: coulomb$

So, charge present in the field is 1000 Coulomb.