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

Byju's Answer

Standard XII

Physics

Electric Field Due to a Sphere and Thin Shell

Two identical...

Question

Two identical positive charges are placed on the y-axis at y = – a and y = + a. The variation of V (potential) along x-axis is shown by graph

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

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

Open in App

Solution

The correct option is A

Potential at centre is $V_{0} = \frac{2 k Q}{a}$

Potential at axis $V_{0} = \frac{2 k Q}{\sqrt{a^{2} + x^{2}}}$

$A t x = \infty V_{\infty} = 0$

Suggest Corrections

Similar questions

Q. Two identical positive charges are placed on the y-axis at y = -a and y = +a. Plot the variation of electric potential along the x-axis.

Q. Two equal positive charges are fixed along x - axis at

- a

and

a

as shown.

E_{x}

and

E_{y}

represent electric field along x - and y- axes respectively and

V

represents potential.

Four identical charges are placed at the four vertices of a square lying in YZ plane with origin at the centre of a
square. A fifth charge is moved along X axis. The variation of potential energy (U) along X axis is correctly
represented by

Q. Two positive equal charges are placed symmetrically about the origin on the x-axis and kept fixed as shown. Another positive charge Q is placed at the origin. The charge Q can be displaced by a very small distance

δ

P_{1}

along the x-axis or to

P_{2}

along the y-axis. Then,

Q. Two positive charges

q_{2} a n d q_{3}

are placed along the y-axis as shown. They exert a net force on

q_{1}

in the positive x- direction. If a third positive charge Q is placed on the x-axis at (X, 0) then the net force on

q_{1}

Join BYJU'S Learning Program

Two identical positive charges are placed on the y-axis at y = – a and y = + a. The variation of V (potential) along x-axis is shown by graph

The correct option is A Potential at centre is V0=2kQa Potential at axis V0=2kQ√a2+x2 At x=∞ V∞=0

The correct option is A

Potential at centre is $V_{0} = \frac{2 k Q}{a}$

Potential at axis $V_{0} = \frac{2 k Q}{\sqrt{a^{2} + x^{2}}}$

$A t x = \infty V_{\infty} = 0$