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gauss law E=Q/(E0*A) Q=793*10^-6 r=1.61 A=4*pi*r^2 E0 = 8.8*10^-12 solve for E
http://www.physicsforums.com/showthread.php?t=209089
Tes
That depends on the strength of the electric field, and on the length of time the electron has been experiencing it. An electron in an electric field accelerates uniformly.
The electric field, in this case, would be the same, no matter how far you go from the plate.
Assuming that the charhe 'q' is uniformly distributed ina sperical volume of radius Discuss the variation of Electric intensity
gauss law E=Q/(E0*A) Q=793*10^-6 r=1.61 A=4*pi*r^2 E0 = 8.8*10^-12 solve for E
Increase the magnitude of the electric current.
Electric field is dependent on the magnitude of the electric charge, E = qzc/r2
http://www.physicsforums.com/showthread.php?t=209089
The magnitude of the electric field is 2.5.
The magnitude of the electric field is 2.5.
Tes
That depends on the strength of the electric field, and on the length of time the electron has been experiencing it. An electron in an electric field accelerates uniformly.
If the charge is uniformly distributed over the shell, then the electric field is zero everywhere inside.
The magnitude of the electric potential is dependent upon the particle's charge and the electric field strength.
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