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 𝐴 and sphere 𝐡 have charges of +π‘žβ‚ and βˆ’π‘žβ‚‚ respectively. The spheres are separated by a straight line distance 𝑑. The net electric field produced by the spheres has a null point along 𝑑 at a distance of 𝑑/4 m from sphere 𝐴. What is the ratio of the magnitude of the charge of sphere 𝐡 to that of sphere 𝐴?

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Video Transcript

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

Let’s start off by drawing a diagram of these two spheres. Our scenario involves two charged spheres, sphere 𝐴 with charge positive π‘ž one and sphere 𝐡 with charge negative π‘ž two. They are separated by a distance 𝑑. And we’re further told that at a point on this line a distance of 𝑑 over four meters from sphere 𝐴, the electric field is zero.

The statement asked us to solve for the ratio of the magnitude of the charges π‘ž two and π‘ž one. We can start figuring that out by recalling the equation for the electric field created by a point charge. A point charge, π‘ž, creates an electric field, 𝐸, equal to the magnitude of the charge times π‘˜, Coulomb’s constant, divided by π‘Ÿ squared, where π‘Ÿ is the distance from the charge to the point where the field 𝐸 is being measured.

Our diagram shows that the charges on sphere 𝐴 and 𝐡 create an electric field of zero at a particular location. If we call that field capital 𝐸, then 𝐸 equals π‘˜ times π‘ž one, the charge on sphere 𝐴, divided by 𝑑 over four quantity squared plus π‘˜ times π‘ž two, the charge on sphere 𝐡, divided by the distance three 𝑑 over four quantity squared.

This sum, we’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 𝑑 over four in our denominators. That leaves us with an equation saying that π‘ž one plus π‘ž two over three squared, or nine, is equal to zero or that π‘ž two equals negative nine π‘ž one.

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

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