Question

Electric field intensity at a point 2m from a point charge is 400N/C. If a charge of 20 micro coulomb is placed at that point, it will experience a force of how much?

Answer 1

Given :

Electric field intensity at a point 2m from a point charge is 400N/C.

To find :

If a charge of 20 micro coulomb is placed at that point, it will experience a force of how much?

Theory :

Electric Field Intensity :

Electric field intensity at a point due to a Given charge is defined as force experienced by a unit positive charge placed at that point.

${\purple{\boxed{\large{\bold{\sf\:\vec{E}=\lim_{q_{o}\to0}\frac{\vec{F}}{q_{o}}}}}}}$

Solution :

Electric field intensity at a point 2m from a point charge is 400N/C

$\implies\:\vec{E}=400\:NC{}^{-1}$

⇒r = distance from point charge =2m

Now a charge of 20μC is placed at that point .

We have to find force experienced by it .

$\bf\:\vec{F}=q\vec{E}$

$\implies\:\sf\:\vec{F}=20\times10{}^{-6}\times400$

$\implies\:\sf\:\vec{F}=20\times10{}^{-6}\times400$

$\implies\:\sf\:\vec{F}=8\times10{}^{-6}\times10{}^{3}$

$\implies\:\sf\:\vec{F}=8\times10{}^{-3}\:N$

Therefore, it will experience $8\times10{}^{-3}\:N$ force.

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More About Electric Field .

1) The SI unit of electric field is N/C or volt/meter .

2)It is a vector quantity and it's direction is the same as force on the positive test charge .

Answer 2

$\mathtt{\huge{ \fbox{Solution :)}}}$

Given ,

Charge (q) = 20 micro C or 20 × (10)^-6 C

Electric field (E) = 400 N/C

We know that , the electrostatic force per unit charge is called electric field

Thus ,

$\sf \hookrightarrow 400 = \frac{f}{20 \times {(10)}^{ - 6} } \\ \\ \sf \hookrightarrow f = 8000 \times {(10)}^{ - 6} \\ \\ \sf \hookrightarrow f = 8 \times {(10)}^{ - 3} \: \: newton$

Hence , the force is 8 × (10)^-3 newton

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