Electric Potential due to two charges

[Mnemonic]

Homework Statement

Two point charges are placed on a horizontal line, the first is +3.0-μC located at x = 0.0-m, y = 0.0-m and the second is -4.0-μC located at x = 2.0-m, y = 0.0-m. At what points on the horizontal line y = 0.0 will the electric potential be zero?

Homework Equations

V(r)=kq/r

The Attempt at a Solution

V(total)=kq1/r1+kq2/r2=0

k(q1/r1+q2/r2)=0

k(3e-6/x-4e-6/(2-x)=0

x=6/7

So there should be no electric potential at x=6/7metres. Is that the only point?

 

[Vibhor]

Homework Statement

Two point charges are placed on a horizontal line, the first is +3.0-μC located at x = 0.0-m, y = 0.0-m and the second is -4.0-μC located at x = 2.0-m, y = 0.0-m. At what points on the horizontal line y = 0.0 will the electric potential be zero?

Homework Equations

V(r)=kq/r

The Attempt at a Solution

V(total)=kq1/r1+kq2/r2=0

k(q1/r1+q2/r2)=0

k(3e-6/x-4e-6/(2-x)=0

x=6/7

So there should be no electric potential at x=6/7metres. Is that the only point?

Your work looks correct . Yes, there is only one point at which potential is zero.

 

[Bystander]

k(q1/r1+q2/r2)=0
k(3e-6/x-4e-6/(2-x)=0

Not quite correct --- you'll want to look at how you've defined "r2."

 

[Mnemonic]

Would I get another valid answer if I first used:

k(3e-6/x-4e-6/(2-x)=0

then used:

k(3e-6/(2-x)-4e-6/x=0
?

 

[Vibhor]

Would I get another valid answer if I first used:

k(3e-6/x-4e-6/(2-x)=0

then used:

k(3e-6/(2-x)-4e-6/x=0
?

No . The answer would remain same i.e point of zero potential will be at a distance (6/7)m from +3.0-μC .

In the latter case ,you are taking 'x' to be the distance of point of zero potential from -4.0-μC .Here you will get x=(8/7)m .Again this point will be at a distance (2-x) i.e (6/7)m from +3.0-μC .

The answer remains same .

 

[Bystander]

You have two charges, opposite in sign, separated by a distance of two meters; at all points on the two meter line segment between those two opposite sign charges there is a non-zero force on any non-zero test charge resulting from the simultaneous attraction and repulsion of the test charge by the two given charges.

 

[Vibhor]

@Mnemonic ,

There is something more you need to do . There might be another solution to this problem.

Consider a point to the left of +3.0-μC at a distance 'x' . Apply the appropriate condition , as you did in the OP .What do you get ?

 

[ehild]

Homework Equations

V(r)=kq/r

The Attempt at a Solution

V(total)=kq1/r1+kq2/r2=0

k(q1/r1+q2/r2)=0

k(3e-6/x-4e-6/(2-x)=0

Remember, r is the distance from the charge, a positive quantity. x is the position along the x axis. If x is positive, 2-x can be both positive and negative, You need to take the absolute value. x can be also negative. So you have to take the absolute values r1=|x|, r2=|2-x|.