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

Byju's Answer

Standard XII

Physics

Vectors and Its Types

Two point cha...

Question

Two point charges $Q$ and $- Q / 4$ placed along the x-axis are separated by a distance $r$ . Take $- Q / 4$ as origin and it is placed at the right of $Q$ . Then the potential is zero.

x = r / 3

only

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

x = - r / 5

only

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

both at

x = r / 3

and at

x = - r / 5

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

there exist two points on the axis where the electric field is zero

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

Open in App

Solution

The correct option is C both at $x = r / 3$ and at $x = - r / 5$

Potential in the right of charge -Q/4= $\frac{- k Q}{4 x_{1}} + \frac{k Q}{r + x_{1}} = 0 o r r + x_{1} = 4 x_{1} o r x_{1} = r / 3$

Potential in the left of charge -Q/4 at distance $x_{2}$ = $\frac{- k Q}{4 x_{2}} + \frac{k Q}{r - x_{2}} = 0 o r x_{2} = \frac{r}{5}$

Suggest Corrections

Similar questions

Q. Two point charges

Q

and

- \frac{Q}{4}

placed along the

x -

axis are seperated by a distance

r

. Take

- \frac{Q}{4}

as origin and it is placed at the right of

Q

. Then the potential is zero

Q. Two point charges

Q

and

- Q / 4

are separated by a distance

x

. Then

Along the X-axis, three charges $\frac{q}{2}, - q$ and $\frac{q}{2}$ are placed at x = 0, x = a and x = 2a respectively. The resultant electric potential at x = a+r (if, a << r) is :

( $ϵ_{0}$ is the permittivity of free space)

Q. Two equal point charges are fixed at

x = - a

and

x = + a

on the

X -

axis Another point charge

Q

is placed at the origin. The change in electrical potential energy

Q

when it is displaced by a small distance

x

along the X -axis, is approximately proportional to

Q. A charge

+ q

is fixed at each of the points

x = x_{0}, x = 3 x_{0}, x = 5 x_{0} . . . . ., \infty

on the x-axis and charge

- q

is fixed at each of the points

x = 3 x_{0}, x = 4 x_{0}, x = 6 x_{0}, . . . . ., \infty

. Here,

x_{0}

is a positive quantity. Take the electric potential at a point due to charge

Q

at a distance

r

from it to be

\frac{Q}{4 π ε_{0} r}

. Then the potential at origin due to the above system of charges is:

Join BYJU'S Learning Program

Two point charges Q and −Q/4 placed along the x-axis are separated by a distance r. Take −Q/4 as origin and it is placed at the right of Q. Then the potential is zero.

The correct option is C both at x=r/3 and at x=−r/5 Potential in the right of charge -Q/4= −kQ4x1+kQr+x1=0orr+x1=4x1orx1=r/3 Potential in the left of charge -Q/4 at distance x2 = −kQ4x2+kQr−x2=0orx2=r5

Two point charges $Q$ and $- Q / 4$ placed along the x-axis are separated by a distance $r$ . Take $- Q / 4$ as origin and it is placed at the right of $Q$ . Then the potential is zero.