Consider the electric force between a pair of charged particles a certain distan

Question

Consider the electric force between a pair of charged particles a certain distance Coulomb's law: If the charge on one of the particles is doubled, the force is If, Instead, the charge on both parties h doubled, the force is If, instead, the distance between the particles is halved, the force is If the distance Is halved, and the charge of both particles is doubled. The force is times as great. In a hydrogen atom, the electron is separated from the proton by an average distance of about 0.53 Times 10^-10 m. Calculate the magnitude of the electrostatic force of attraction exerted by the proton on the electron. Refer to the figure below. Which charge is positive? Which charge has a larger magnitude?

Explanation / Answer

formula for force between the charges is given by F = Kq1q2/r^2

part A : if one q doubled, F also doubles , option C

part B: if both charges doubled, F becomes 4 times , option D

part C: if d is reduced to half, F = Kq1q2/d/2^2 = 4 times of F option D

part D: if d is halved and both charges doubled, F = 4 kq1q2/d/2^2 = 16 times great

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7. F = Kq1q2/r^2

q1 = 1.6 e -19 C

q2 = 1.6 e -19 C

r = 0.53 e -10 m

F = 9e9 * 1.6 e -19 * 1.6 e-19/(0.53 e -10 * 0.53 e-10)

F = 8.2 e -8 N
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8.

for positive charge field lines are outward and for -ve charge field lines are inward

so

q1 is +Ve

more field lines correspond to more charge

so q1 is larger magnetude

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