A uniform electric field of 500 N/C is directed in the -x direction. How does th

Question

A uniform electric field of 500 N/C is directed in the -x direction. How does the electric potential at point A (4 m, 3 m) compare to the electric potential at point B (3 m, 3 m)? The potential at point A is larger The potential at point B is larger The potential at points A and B are the same. A charge of -3 mu C is located at the origin. What is the force on the charge? If the charge is moved from the origin to the point (4 m, 0), what is its change in electrostatic potential energy? Two charged spheres are a distance R apart and exert a force F on each other. If the distance between the two charges is tripled, what is the force between the charges? F 3F 9F F/3 F/9

Explanation / Answer

1) Given :-

Electric field = 500N/C

Electric potential at point A(4m,3m),

R = sqrt(x^2 + y^2) where x= 4m, y=3m R is resultant distance

= sqrt(16+9) = 5m

Now electric potential at A =E.R = 500*5= 2500V ------------------------------------(i)

Now at point B (3m,3m)

R = sqrt(9+9)

R = 4.242

Electric potential at point B = E.R = 4.242* 500 = 2121.320 V --------------------(ii)

So electric potentail at point A is greater than point B--------------------------------------------Answer

B)As we know ,

E= F/Q

F = E.Q = 500*-3*10^-6 = -0.0015N ---------------------------Answer

C) Change in energy can be calculated as Q*d*E where d is the distance moved by a charge Q

Q=-3* 10^-6 d= 6m,E= 500n/c

= -3*10^-6*4*500 J =6mJ----------------------------------------------Answer

4) As we know that, from Columb law

F=k(q1q2)/ d2

force is directly propotional to charge and inversaly propotional to square od distance

so as per the question if we incread distance by 3 then F will decreased by 9 times

hence F/9 will be the answer

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