Answer:
E = 12262 V/m
Explanation:
given,
Charge of the particle, q = +4.0 μC
mass of the ball, m = 5 g
Electric field, E = ?
Force due to weight
F = m g
Force due to electric field
F =q E
To balance the weight of particle both the forces must be equal
Electric field exerted on the ball will be equal to
m g = q E
E = 12262 V/m
Hence, the electric field acting in the particle is equal to E = 12262 V/m
Answer:
Explanation:
A body has acceleration when there is a change in the velocity vector, either in magnitude or direction. In this case we only have a change in magnitude. The average acceleration represents the speed variation that takes place in a given time interval.
a)
b)
The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m
Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:
Substitute numerical values:
The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
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