Two point charges 2C and −1C are placed at x=0 and x=6 respectively. The potential will be zero at points :
x=2,−2
x=1,5
x=4,12
x=2,9
A
x=2,9
B
x=2,−2
C
x=1,5
D
x=4,12
Open in App
Solution
Verified by Toppr
Since the two charges are opposite, in between them there will be a point where the potential is zero
Let the point be at distance a from x=0.
We know thatV=14πϵ0r where r is the distance from the charge and the point. The total potential will be the sum of the two potentials.
From the conditions of the problem, we have
24πϵ0a−14πϵ0(6−a)=0
here we get a=4
Also, there will be a point on the right side of the -1C charge where potential = 0. Let that point be at a distance b from x=0.
From the conditions of the problem, we have
24πϵ0b−14πϵ0(b−6)=0
Here we get b=12
Was this answer helpful?
1
Similar Questions
Q1
Two point charges 2C and −1C are placed at x=0 and x=6 respectively. The potential will be zero at points :
View Solution
Q2
Three point charges Q,q and Q are placed along a straight line placed along X axis at point x=0,x=L2 and x=L respectively. The entire system, of three charges will be in equilibrium, if
View Solution
Q3
An electric charge 10−3μC is placed at the origin (0, 0) of X-Y co-ordinate system. Two points A and B are situated at (√2,√2) and (2, 0) respectively. The potential difference between the points A and B will be :
View Solution
Q4
Charges of q coulomb but of alternately opposite signs are placed along the x-axis at x=1, x=2, x=4, ... and so on. The electric potential at the point x=0 due to all these charges will be
View Solution
Q5
Point charges q1 and q2 lie on the x−axis at points x=−a and x=+a respectively. Given the net electrostatic force on point charge +Q placed at x=+a/2 to be zero and q1=nq2. Find n.