The force acting on a charge of $10^{- 10} C$ placed in an electric field of intensity $600$ V/m is :

6 \times 10^{12} N

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6 \times 10^{- 8} N

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6 \times 10^{- 12} N

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6 \times 10^{8} N

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Solution

The correct option is A $6 \times 10^{- 8} N$
Given charge: $q = 10^{- 10} C$

$E = 600 v / m$

Force $= q . E$
$= 600 \times 10^{- 10} N$
$= 6 \times 10^{- 8} N$

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The force acting on a charge of 10−10C placed in an electric field of intensity 600 V/m is :

The correct option is A 6×10−8NGiven charge: q=10−10CE=600 v/mForce =q.E =600×10−10N =6×10−8N

The force acting on a charge of $10^{- 10} C$ placed in an electric field of intensity $600$ V/m is :

The correct option is A $6 \times 10^{- 8} N$
Given charge: $q = 10^{- 10} C$

$E = 600 v / m$

Force $= q . E$
$= 600 \times 10^{- 10} N$
$= 6 \times 10^{- 8} N$