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Q.The unit of intensity of electric field is –

(a)
(i) newton/metre
(b)
(ii) newton/coulomb
(c)
(iii) joule/newton
(d)
(iv) coulomb/newton
Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2018MCQ· 1mImportance★★★★★
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The correct option is (ii) newton/coulomb (N/C), because electric field intensity is force per unit charge.

Concept. The electric field intensity EE at a point is the electrostatic force FF that would act on a small positive test charge qq placed there, divided by that charge:

E=FqE = \frac{F}{q}

Why this gives the unit. Substituting SI units of the quantities on the right:

[E]=[force][charge]=newtoncoulomb=N C−1[E] = \frac{[\text{force}]}{[\text{charge}]} = \frac{\text{newton}}{\text{coulomb}} = \text{N C}^{-1}

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