Question : Two charges -q and +q are located at points (0,0,-a) and (0,0,a) respectively.
i) What is the electrostatic potential at the points (0,0,z) and (x,y,0)?
ii) How much work is done in moving a small test charge from the point (5,0,0) to (-7,0,0) along the x-axis?
Doubt by Saif
Solution :
Two charges -q and +q are located at points (0,0,-a) and (0,0,a) respectively. We can assume that an electric dipole is placed along the z-axis and it centre is at the origin. The dipole is having charges -q and +q and it is separated by a distance 2a
i) The coordinate (0,0,z) lies on the positive z axis and the dipole is along lies along the z axis. So, the electric potential at (0,0,z) will be equal to the potential at an axial point of dipole and its magnitude will be given by
Vaxial = kp/(z²-a²) where k = 1/4πε₀
All the point which lie in a XY Plane will be perpendicular to the z-axis. Since the dipole is lying on z-axis so any point (x,y,0) will lie on the equatorial plane of the dipole and we know, electric potential at any equatorial plane/point of the dipole is zero i.e. Vequ = 0.
ii) Let A(5,0,0) & B(-7,0,0)
Both the point A and B lies on the equatorial axis of the dipole so we can say that VA=VB=0
Work done = charge × Potential difference between the two points.
W = q₀×(VB-VA)
W = q₀×0
W = 0
Hence, there would be zero work done in moving a small test charge from the point (5,0,0) to (-7,0,0) along the x-axis.
