# Question Video: Calculating the Charge Required to Produce a Particular Absolute Electric Potential

What is the magnitude of a point charge that produces a potential of 7.47 V at a distance of 2.63 mm?

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### Video Transcript

What is the magnitude of a point charge that produces a potential of 7.47 volts at a distance of 2.63 millimeters?

We can imagine this point charge then located at some spot in space, and we want to solve for the magnitude of that charge. We’ll call it 𝑄. We’re told that at a distance from 𝑄 of 2.63 millimeters this charge produces an electric potential of 7.47 volts.

Based on all this information, we want to work backwards to solve for the charge 𝑄. To start off, let’s recall the mathematical relationship that ties together charge, electric potential, and distance. For a point charge 𝑄, the electric potential created by that charge is equal to 𝑘, Coulomb’s constant, times 𝑄 divided by the distance from the charge to the point at which the potential is being calculated.

And we know that this formulation assumes that electric potential is zero infinitely far away from this point charge. 𝑘, that Coulomb’s constant we mentioned, can be approximated as 8.99 times 10 to the ninth newtons meter squared per coulomb squared. Knowing all this, we have a relationship that brings together the quantities we have and the quantity we want to solve for.

That quantity is 𝑄, so let’s rearrange this expression to isolate it on one side. The point charge magnitude 𝑄 equals the potential 𝑉 multiplied by 𝑟, the distance between 𝑄 and where 𝑉 is measured, all divided by Coulomb’s constant. We’re just about ready to plug into this expression and solve for 𝑄. But one last thing, before we do, we want to change the units of our distance from millimeters into units of meters.

We’ll do that so that those units are consistent with those in the rest of the expression. If we recall that 1000 millimeters is equal to one meter, that means we’ll want to move the decimal place in our millimeters value three spots to the left to convert it to a value in meters. So 2.63 millimeters is equal to 0.00263 meters.

This is the value we’ll use for 𝑟 in our expression to solve for 𝑄. When we plug in for these values and then calculate 𝑄, to three significant figures, it comes out to a value of 2.19 times 10 to the negative twelfth coulombs of charge. That’s the magnitude of the charge that produces the given potential at the given distance.