Two point charges both 1 mu C are separated by 2 cm as shown in box A. What is t

Question

Two point charges both 1 mu C are separated by 2 cm as shown in box A. What is the resultant electric potential v at the point P that lies 3 cm below the midpoint of the two charges in box A (taking V = 0 at infinity)? If the two changes are replaced with 2 cm-long uniformly charged line with the same total charge (-2 mu C).with midpoint 3 cm above P (as shown in box B), will the magnitude of the electric field at P be larger, smaller, or the same as it was in box A? Explain your reasoning What is the electric field at point P in box A (express your answer in vector notation) d) An electron s moving with velocity S cm/s in the x direction in an electric field E = 3 V/cm that points in the -y direction What is the direction and magnitude the electron's acceleration?

Explanation / Answer

a)

distance of P from each source r = sqrt(1+9) = 3.16 cm

potential at P ( the magnitude of the potential is ame form each as both are of same sign)

      Vp = 2*kq/r = -2*9.0e+9*1.0e-6/3.16e-2 = -5.7e+5 V

b) If the two sources are replaced by a uniformly charged line siurce then the field at P would be larger than that in Box A as some of the elemnts on the line source are closer to P than when all the charge is concentrated at the ends. Field is inversly proportional to distance.

c) Field at P due to eahc charge

   E = kq/r2 = 9.0e+9*1.0e-6 /1.0e-3

                     = 9.0e+6 N/c

The direction of each of the field will be along the line joning the charge to P.

resolve E parallel to the line and perpendicular, the paralle components are equal in magnitude but opposite in direction and hence cancel up

The verticle componnets will addup

The net field En = 2* 9.0e+6 (3/3.16)

                            = 17.09 j  N/C ( taking +y-axis along P joining the mid point of the charges )

d) The electron will experince a force of qE in the -y direction

                           field E = 3 V/cm   = 300 V/m   = 300 N/C

Foce on the electron = -1.6e-19 *300 = -4.8e-17 N

acceleration a = F/m   = 4.8e-17/9.0e-31

                          = -5.3 e-13 m/s2

the electron is accelerated in +y-direction

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