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Question: A force of 2.25 N acts on a charge of \(15 \times 10^{- 4}C\).The intensity of electric field at tha...

A force of 2.25 N acts on a charge of 15×10−4C15 \times 10^{- 4}C.The intensity of electric field at that point is.

A

150 N C−1C^{- 1}

B

15 N C−1C^{- 1}

C

1500 N C−1C^{- 1}

D

1.5 N C−1C^{- 1}

Answer

1500 N C−1C^{- 1}

Explanation

Solution

: Electric field.

E=Fq=2.25N15×10−4C=1500NC−1E = \frac{F}{q} = \frac{2.25N}{15 \times 10^{- 4}C} = 1500NC^{- 1}