Two particles, X and Y. are 4 m apart. X has a charge of 2Q and Y has a charge o

Question

Two particles, X and Y. are 4 m apart. X has a charge of 2Q and Y has a charge of Q. The force of X on Y has twice the magnitude of the force of Y on X has half the magnitude of the force of Y on X has four times the magnitude of the force of Y on X has one-fourth the magnitude of the force of Y on X has the same magnitude as the force of Y on X Two protons (p1 and p2) and an electron (e) lie on a straight line, as shown. The directions of the force of p1 on e. the force of p2 on e, and the total force on e, respectively, are. rightarrow, leftarrow, rightarrow leftarrow, leftarrow, leftarrow rightarrow, leftarrow, leftarrow leftarrow, leftarrow, leftarrow, leftarrow, leftarrow, leftarrow

Explanation / Answer

4 ) The electric foce by one charge on another is given by F = k*Q1*Q2/r2 . Hence The force on Q1 by Q2 will be

F1 =  k*Q1*Q2/r2 = k*2q*q/42 = kq2/8, and force on Q2 by Q1 will be

F1 =  k*Q2*Q1/r2 = k*q*2q/42 = kq2/8 Hence |F1| = |F2|. Force between two charges will always equal in magnitude and opposite in direction. hence option(e) is correct.

5) Since proton and electron have opposite charges hence both protons will attract the electron. Hence p1 will attract electron towards left (<--) and p2 towards right (-->). Since p1 is closer to electron hence force due to p1 will be larger than p2, which means net force will be towards left(<--). Hence correct option will be (d) <--, --> , <--

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