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Byju's Answer

Standard XII

Physics

Electric Field Due to a Sphere and Thin Shell

A positively ...

Question

A positively charged ball hangs from a long silk thread. Electric field at a certain point (at the same horizontal level of ball) due to this charge is E. Let us put a positive test charge $q_{0}$ at this point and measure F/ $q_{0}$ on this charge. Then,

E > F / q_{0}

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E < F / q_{0}

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E = F / q_{0}

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none of these

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Solution

The correct option is A $E > F / q_{0}$
$E = \frac{k Q}{r^{2}}$

$F = \frac{k Q q_{0}}{r_{1}^{2}} o r \frac{F}{q_{0}} = \frac{k Q}{r_{1}^{2}}$ (ii)

As $r_{1} > r$ , so from equation (i) and (ii), we get $E > F / q_{0}$

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A positively charged ball hangs from a long silk thread. Electric field at a certain point (at the same horizontal level of ball) due to this charge is E. Let us put a positive test charge q0 at this point and measure F/q0 on this charge. Then,

The correct option is A E>F/q0E=kQr2F=kQq0r21orFq0=kQr21 (ii)As r1>r, so from equation (i) and (ii), we get E>F/q0

A positively charged ball hangs from a long silk thread. Electric field at a certain point (at the same horizontal level of ball) due to this charge is E. Let us put a positive test charge $q_{0}$ at this point and measure F/ $q_{0}$ on this charge. Then,

The correct option is A $E > F / q_{0}$
$E = \frac{k Q}{r^{2}}$

$F = \frac{k Q q_{0}}{r_{1}^{2}} o r \frac{F}{q_{0}} = \frac{k Q}{r_{1}^{2}}$ (ii)

As $r_{1} > r$ , so from equation (i) and (ii), we get $E > F / q_{0}$