# Question Video: Finding the Ratio of the Charges on Two Objects from the Position of the Null Point of the Electric Field between Them

Two equal-sized charged spheres, sphere π΄ and sphere π΅ have charges of +πβ and βπβ respectively. The spheres are separated by a straight line distance π. The net electric field produced by the spheres has a null point along π at a distance of π/4 m from sphere π΄. What is the ratio of the magnitude of the charge of sphere π΅ to that of sphere π΄?

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### Video Transcript

Two equal sized charged spheres, sphere π΄ and sphere π΅, have charges of plus π one and negative π two respectively. The spheres are separated by a straight-line distance π. The net electric field produced by the spheres has a null point along π at a distance of π over four meters from sphere π΄. What is the ratio of the magnitude of the charge of sphere π΅ to that of sphere π΄?

Letβs start off by drawing a diagram of these two spheres. Our scenario involves two charged spheres, sphere π΄ with charge positive π one and sphere π΅ with charge negative π two. They are separated by a distance π. And weβre further told that at a point on this line a distance of π over four meters from sphere π΄, the electric field is zero.

The statement asked us to solve for the ratio of the magnitude of the charges π two and π one. We can start figuring that out by recalling the equation for the electric field created by a point charge. A point charge, π, creates an electric field, πΈ, equal to the magnitude of the charge times π, Coulombβs constant, divided by π squared, where π is the distance from the charge to the point where the field πΈ is being measured.

Our diagram shows that the charges on sphere π΄ and π΅ create an electric field of zero at a particular location. If we call that field capital πΈ, then πΈ equals π times π one, the charge on sphere π΄, divided by π over four quantity squared plus π times π two, the charge on sphere π΅, divided by the distance three π over four quantity squared.

This sum, weβre told, is equal to zero. Considering this equation, we see that we can cancel out both the factors of π and the factors of π over four in our denominators. That leaves us with an equation saying that π one plus π two over three squared, or nine, is equal to zero or that π two equals negative nine π one.

Since our problem statement specifically asked for the magnitude of the ratio of π two to π one, we can remove the minus sign we see here. And we see that the magnitude of charge π two equals nine times the magnitude of charge π one. Thatβs the charge magnitude ratio that creates this electric field null point.