Two metal plates having a potential difference of 800 V and 2 cm apart. It is found that a particle of mass 1.96 × 10-15 kg remaining suspended in the region between the plates. The charge on the narticle must be (e= elementry charge)

Question

Two metal plates having a potential difference of 800 V and 2 cm apart. It is found that a particle of mass 1.96 × 10-15 kg remaining suspended in the region between the plates. The charge on the narticle must be (e= elementry charge) (A) 8e (B) 6e (C) 3e (D) 2e

Tardigrade Admin · Accepted Answer

E=(V/d)=(800/2 × 10-2) V / m =4 × 104 V / m again, q E=m g q=(m g/E)=(1.96 × 10-15 × 9.8/4 × 104) e =3 e

Q. Two metal plates having a potential difference of $800 V$ and $2 c m$ apart. It is found that a particle of mass $1.96 \times 1 0^{- 15} k g$ remaining suspended in the region between the plates. The charge on the narticle must be $(e =$ elementry charge)

8e

6e

3e

2e

$E = \frac{V}{d} = \frac{800}{2 \times 1 0 ^{- 2}} V / m$
$= 4 \times 1 0^{4} V / m$
again, $qE = m g$
$q = \frac{m g}{E} = \frac{1.96 \times 1 0 ^{- 15} \times 9.8}{4 \times 1 0 ^{4}} e$
$= 3 e$

Q. Two metal plates having a potential difference of 800V and 2cm apart. It is found that a particle of mass 1.96×10−15kg remaining suspended in the region between the plates. The charge on the narticle must be (e= elementry charge)

8e

6e

3e

2e

E=dV​=2×10−2800​V/m =4×104V/m again, qE=mg q=Emg​=4×1041.96×10−15×9.8​e =3e

Q. Two metal plates having a potential difference of $800 V$ and $2 c m$ apart. It is found that a particle of mass $1.96 \times 1 0^{- 15} k g$ remaining suspended in the region between the plates. The charge on the narticle must be $(e =$ elementry charge)

$E = \frac{V}{d} = \frac{800}{2 \times 1 0 ^{- 2}} V / m$
$= 4 \times 1 0^{4} V / m$
again, $qE = m g$
$q = \frac{m g}{E} = \frac{1.96 \times 1 0 ^{- 15} \times 9.8}{4 \times 1 0 ^{4}} e$
$= 3 e$