Q. 594.0( 5 Votes )
An electric field exists in the space, where A = 10 Vm–2. Take the potential at (10m, 20m) to be zero. Find the potential at the origin.
Electric Field :
A = 10 Vm-2
V(10m,20m) = 0
Change in potential is:
Where dV is change in potential, E is the electric field and dx is the change in displacement.
In vector form:
Integrating we get:
Here limits 10 to 0 is taken as 10 m is the x co-ordinate and origin has (0,0)
Hence. potential of 500 V exists at the origin.
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(a) In a quark model of elementary particles, a neutron is made of one up quarks [charge (2/3) e] and two down quarks [charges – (1/3) e]. Assume that they have a triangle configuration with side length of the order of 10–15 m. Calculate electrostatic potential energy of neutron and compare it with its mass 939 MeV.
(b) Repeat above exercise for a proton which is made of two up and one down quark.
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