Two small charged objects repel each other with a force of magnitude F when sepa

Question

Two small charged objects repel each other with a force of magnitude F when separated by a distance r. If the charge on each object is reduced to one-fourth of its original value and the distance between them is reduced to one-half of its original value, the magnitude of the new force becomes: F/16 F/8 F/4 F/2 F Two particles, X and Y, are 4 m apart. X has a charge of 2Q and Y has a charge of Q. The ratio of the electrostatic force on particle X to that of the electrostatic force on particle Y is: 4:1 2:1 1:1 1:2 1:4 Charges +Q, -Q. and q are placed at the vertices of an equilateral triangle, as shown. No that | + Q| = | - Q|. The net force exerted on the charge q is: toward charge + Q toward charge - Q away from charge +Q perpendicular to the line joining +Q and -Q parallel to the line joining + Q and -Q

Explanation / Answer

here,

the magnitude of force is F

when the distance between the objects is r

when the disatnce is halved , r' = r/2

and charge , Q' = Q/4

the magnitude of force , F' = G * Q'^2 /r'^2

F' = G * ( Q/4)^2 /( r/2)^2

F' = F /4

the magnitude of force is C) F/4

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