First slide

Electric field

Question

The intensity of electric field required to balance a proton of mass $1 .7 \times 10^{- 27} kg$ and charge $1 .6 \times 10^{- 19} C$ is nearly

Easy

A

$1 \times 10^{- 7} V / m$
B

$1 \times 10^{- 5} V / m$
C

$1 \times 10^{7} V / m$
D

$1 \times 10^{5} V / m$

Solution

Since $qE = mg or E = \frac{mg}{q} = \frac{1 .7 \times 10^{- 27} \times 9 .8}{1 .6 \times 10^{- 19}}$
$= 10 .0 \times 10^{- 8} = 1 \times 10^{- 7} V / m$

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