A charged particle of mass 0.003 g is held stationary in space by placing it in a downward direction of electric field of 6 × 104 N/C. Then the magnitude of charge is

Question

A charged particle of mass 0.003 g is held stationary in space by placing it in a downward direction of electric field of 6 × 104 N/C. Then the magnitude of charge is (A)  5× 10 - 4 C (B)  5× 10 - 10 C (C)  5× 10 - 6 C (D)  5× 10 - 9 C

Tardigrade Admin · Accepted Answer

By using qE = mg q = ( mg / E ) = ( 3 × 10 - 6 × 10/ 6 × 104) = ( 30 × 10 10/ 6 ) = 5 × 10 - 10 C

Q. A charged particle of mass 0.003 g is held stationary in space by placing it in a downward direction of electric field of $6 \times 1 0^{4}$ N/C. Then the magnitude of charge is

$5 \times 1 0^{- 4}$ C

$5 \times 1 0^{- 10}$ C

$5 \times 1 0^{- 6}$ C

$5 \times 1 0^{- 9}$ C

By using qE = mg
q = $\frac{m g}{E}$
= $\frac{3 \times 1 0 ^{- 6} \times 10}{6 \times 1 0 ^{4}} = \frac{30 \times 1 0 ^{10}}{6}$
= $5 \times 1 0^{- 10}$ C

Q. A charged particle of mass 0.003 g is held stationary in space by placing it in a downward direction of electric field of 6×104 N/C. Then the magnitude of charge is

5×10−4 C

5×10−10 C

5×10−6 C

5×10−9 C