You visited us 1 times! Enjoying our articles? Unlock Full Access!

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}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

E < F / q_{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

E = F / q_{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $E > F / q_{0}$

Suggest Corrections

Similar questions

Q. A positively charged ball hangs from a silk thread. We put a positive test charge

q_{0}

at a point and measure

F / q_{0}

, then it can be predicted that the electric field strength E.

Q. A positively charged ball hangs from a long silk thread. We put a positive test charge q

_{0}

at a point and measure F/q

_{0}

, then it can be predicted that field E

If a charged body experiences a force of repulsion of magnitude $F$ from a charge of $q_{0}$ , the electric field $E$ of the charged body at a point lying not in between the $2$ charges but on the same line joining two charges, is :

Q. Two equal positive point charges

Q

are fixed at points

B (a, 0)

and

A (- a, 0)

. Another test charge

q_{0}

is placed at

O (0, 0)

. Then select the most appropriate statement regarding the equilibrium position of

q_{0}

for displacement along

x -

axis.

Q. A thin insulating ring of radius

R

has a uniform linear charge density

λ

. Now a charge

Q_{0}

is placed at centre of this ring. The increment in the stretching force of the wire due to the charge

Q_{0}

is:-

Join BYJU'S Learning Program

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 $E > F / q_{0}$