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 the ball is E . If now put a positive test charge q0 at this point and measure (F/q0) , here F is the force on q0 due to the charged ball. Then E is:

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 the ball is E . If now put a positive test charge q0 at this point and measure (F/q0) , here F is the force on q0 due to the charged ball. Then E is: (A) >(F/q0) (B)  < (F/q0) (C) =(F/q0) (D) none of these

Tardigrade Admin · Accepted Answer

E=(k Q/r2) .....(i) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-ei8ffptuwy9ysj4k.jpg" /> F=(k Q q0/r12) ⇒ (F/q0)=(k Q/r12) ....(ii) As r1>r, so from equation (i) and (ii), we get E>(F/q0)

$> \frac{F}{q _{0}}$

$< \frac{F}{q _{0}}$

$= \frac{F}{q _{0}}$

none of these

$E = \frac{k Q}{r ^{2}}$ .....(i)

$F = \frac{k Q q _{0}}{r _{1}^{2}} \Rightarrow \frac{F}{q _{0}} = \frac{k Q}{r _{1}^{2}}$ ....(ii)
As $r_{1} > r,$ so from equation (i) and (ii), we get $E > \frac{F}{q _{0}}$

Q. 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 the ball is E . If now put a positive test charge q0​ at this point and measure q0​F​ , here F is the force on q0​ due to the charged ball. Then E is:

>q0​F​

<q0​F​

=q0​F​

none of these

E=r2kQ​ .....(i) F=r12​kQq0​​⇒q0​F​=r12​kQ​ ....(ii) As r1​>r, so from equation (i) and (ii), we get E>q0​F​

$> \frac{F}{q _{0}}$

$< \frac{F}{q _{0}}$

$= \frac{F}{q _{0}}$

$E = \frac{k Q}{r ^{2}}$ .....(i)

$F = \frac{k Q q _{0}}{r _{1}^{2}} \Rightarrow \frac{F}{q _{0}} = \frac{k Q}{r _{1}^{2}}$ ....(ii)
As $r_{1} > r,$ so from equation (i) and (ii), we get $E > \frac{F}{q _{0}}$