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Problem 65 Hard Difficulty

Two hard rubber spheres, each of mass $m=15.0 \mathrm{g}$ are rubbed with fur on a dry day and are then suspended with two insulating strings of length $L=$ 5.00 $\mathrm{cm}$ whose support points are a distance $d=$ 300 $\mathrm{cm}$ , from each other as shown in Figure $P 15.65$ . During the rubbing process, one sphere receives exactly twice the charge of the other. They are observed to hang at equilibrium, cach at an angle of $\theta=10.0^{\circ}$ with the vertical. Find the amount of charge on cach spherc.

Answer

Therefore, the charge that remains on both the spheres is $5.69 \times 10^{-8} \mathrm{C}$ and $1.14 \times 10^{-7} \mathrm{C}$ .

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Video Transcript

in this problem. On the topic of electric forces and fields, we are given to rubble spheres in the figure each with the same mass, which is given and being suspended by two insulating strings of a given length. We know the distance between the support points of each of these strings, and we are told that during the rubbing process, once fear receives twice the charge of the other, so they observed to hang an equilibrium at an angle of 10 degrees with the vertical. We want to find the amount of charge on each of these fears. So if we draw a free body diagram for the charge on the right, we can see the tension force, which acts of 10 degrees to the vertical, the electric force, which is repulsive, as well as the weight acting vertically downward. Now the two spheres have charges. Q one and Q two to Q two is two times key one. So the repulsive force that one exists and the other has the magnitude. F E is equal to K e. Q. One. You too, over r squared, which we can write as to K e e. Q. One squared over R squared. Now we can observe that the distance separating the two spheres is our and are we can see is D, which is three centimeters plus two into the length of the string, five centimeters times the sine of 10 degrees. And so we get our to be four 0.74 centimeters or 0.0 474 meters. Now from the force diagram of one sphere given on the right, we can see that be some of the forces in the Y Direction is equal to zero due to equilibrium, meaning that the Y component of the tension t costs 10 degrees must balance the weight of the sphere mg or simply detention. T is equal to mg over the call sign of 10 degrees. Similarly, the forces in the X direction must add to zero, which means that the X component of the tension T sign 10 degrees is equal to mg over the co sign of 10 degrees times the sign of 10 degrees, which is MG 10 10 degrees. So therefore we can see that the Force two K e Q one squared over R squared is equal to mg 10 of 10 degrees. So then, from here we can find Q one we find Q one. We can find you two so Q one is equal to the square root of MG R squared 10 of 10 degrees divided by two times the electric constant K e. Since we know these values, we can put them into the situation and solve for Q one. So this is the mess of one sphere 0.0 15 kg times G, which is 9.8 m per square. Second times are squared, which is zero point zero 474 meters squared times the 10 of 10 degrees. This is all divided by two times the constant 8.99 times 10 to the power of Maine Newton meters squared Coolum squared. So calculating we get Q one to be five 0.69 times 10 to the minus eight columns. And since we have Q one, we can now find you two. We know Q two is two times key one, So this gives us one 0.14 times 10 to the minus seven columns, so we have the charge of each of the two spheres

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Cornell University

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