# Electric Potential due to two charges

• Mnemonic
In summary, the electric potential will be zero at the point x=6/7m from the +3.0-μC charge and at the point x=8/7m from the -4.0-μC charge. The electric potential will also be zero at any point on the line between these two points.
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?

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?

Last edited:
Mnemonic said:

## 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?

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.

Mnemonic
Mnemonic said:
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."

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
?

Mnemonic said:
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 .

Mnemonic
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.

@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 ?

Mnemonic
Mnemonic said:

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|.

## 1. What is electric potential?

Electric potential is the amount of work needed to move a unit of positive charge from a reference point to a specific point in an electric field. It is measured in volts (V).

## 2. How is electric potential due to two charges calculated?

The electric potential due to two charges is calculated by adding the electric potentials individually. This can be done using the formula V = k(q1/r1 + q2/r2), where k is the Coulomb's constant, q1 and q2 are the charges, and r1 and r2 are the distances from the charges to the point where the electric potential is being measured.

## 3. What is the difference between electric potential and electric field?

Electric potential is a measure of the potential energy of a charge in an electric field, while electric field is a measure of the force experienced by a charge in an electric field. In other words, electric potential is a scalar quantity, while electric field is a vector quantity.

## 4. What is the relationship between electric potential and distance from a charge?

The electric potential is directly proportional to the distance from a charge. This means that as the distance increases, the electric potential decreases. This relationship is described by the inverse square law, which states that the electric potential is inversely proportional to the square of the distance.

## 5. How does the direction of the electric potential due to two charges depend on the relative positions of the charges?

The direction of the electric potential due to two charges depends on the relative positions of the charges. If the charges have the same sign, the electric potential is positive and points away from the charges. If the charges have opposite signs, the electric potential is negative and points towards the charges. The direction can also be determined using the electric field lines, which always point in the direction of decreasing electric potential.

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