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Electric Field Due to a Line of Charge Not on Its Axis

A positively ...

Question

A positively charged ball hangs from a long silk thread. The 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, $E$

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 $E > F / q_{0}$
$E = \frac{k Q}{r^{2}} . . . . (i)$
$F = \frac{k Q q_{0}}{r_{1}^{2}} \Rightarrow \frac{F}{q_{0}} = \frac{k Q}{r_{1}^{2}} . . . (i i)$
As $r_{1} > r$ so equations (i) and (ii) we get $E > F / q_{0}$

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A positively charged ball hangs from a long silk thread. The 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, E

The correct option is A E>F/q0E=kQr2....(i)F=kQq0r21⇒Fq0=kQr21...(ii)As r1>r so equations (i) and (ii) we get E>F/q0

A positively charged ball hangs from a long silk thread. The 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, $E$

The correct option is A $E > F / q_{0}$
$E = \frac{k Q}{r^{2}} . . . . (i)$
$F = \frac{k Q q_{0}}{r_{1}^{2}} \Rightarrow \frac{F}{q_{0}} = \frac{k Q}{r_{1}^{2}} . . . (i i)$
As $r_{1} > r$ so equations (i) and (ii) we get $E > F / q_{0}$