1.

2.25 N force is acting on 15 xx 10^(-4)C charge placed at a point in a uniform electric field. So, intensity of this electric field is......

Answer»

150 N
15 N
`1500 N//C`
`0.15 N//C`

SOLUTION :`q = 15 xx 10^(-4) C, F= 2.25 N, E=?`
`THEREFORE E = F/q`
`=(2.25)/(15 xx 10^(-4))`
`=1500 V//m`


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