This star has a ball with a positive charge Q at each point. Use Q= 2.50 10-6 C,

Question

This star has a ball with a positive charge Q at each point. Use Q= 2.50 10-6 C, and assume that each ball Is a distance of 60.0 cm away from the exact center of the star. Also, use k= 9.00 109 N.m2/C2. If e place a sixth charged ball at the exact center of the star, with a charge of +2Q, what is the magnitude of the net electrostatic force exerted an it by the five charged balls shown above? As shown below, we now replace one of the balls of charge Q (the one that was at the top point of the star) by a ball of charge=3Q. Once again, we want to know the magnitude of the net electrostatic force exerted on a ball of charge +2Q located at the exact center of the star, by the five charged balls shown below. One method to do this is to work out the five individual forces, and add them as vectors, but we should be able to do less work by trying another method, as outlined in parts (b) and A simpler method is to make use of the answer to part (a). What do you add to the first picture to get the second picture? (Enter a sign and number only.) The second picture above, with the ball of -3Q, can be thought of as the first picture (the one with five identical charged balls) plus a single charged ball, added at the top point of the star, that has a charge of Looked at from the perspective described in (b), the net force an a ball of charge +2Q, located at the exact center of the star in the second picture, is the vector sum of the answer to part (a), plus the force exerted on the ball of charge +2Q by the ball that is the answer to part (b). This gives a net force, acting on a ball of charge +2 the exact center of the star that has a magnitude of N. The direction of the net force in part (c) has components that are directed where Select all that apply. If the force has components directed down and to the left, for instance, you would select both down" and left.") up down left right

Explanation / Answer

A) Force = 0 N because electric field was zero because of symmetry

B) - 4Q

to reach the given configuration we added - 4Q at top corner to initiate no-electric field configuration.

C) Force = k*4Q*2Q/r^2 = 9e9*8*2.5e-6^2 /0.60^2

= 1.25 N answer

D) up because it is attractive force

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