logo
or
search
upload

Two charges -q and +q are located at points (0, 0, –a) and (0, 0, a), respectively. <br> (a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0) ? (b) Obtain the dependence of potential on the distance r of a point from the origin when r//a gt gt 1. <br> (c) 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? Does the answer change if the path of the test charge between the same points is not along the x-axis?

Apne doubts clear karein ab Whatsapp par bhi. Try it now.
CLICK HERE

Loading DoubtNut Solution for you

Watch 1000+ concepts & tricky questions explained!

115.6 K+ views | 5.8 K+ people like this

   
Share
Share

Answer Text

Solution :
(a) zero at both the points <br> Charge -q is located at (0,0,-a) and charge +q is located at (0,0,a) . Hence they from a dipole. Point (0,0,z) is on the axis of this dipole and point (x,y,0) is normal to the axis of the dipole . Hence, electrostatic potential `(V_(1)) ` at point (5,0,0) is given by <br> `V_(1)=(-q)/(4piin_(0))1/(sqrt((5-0)^(2)+(-a)^(2)))+q/(4piin_(0))1/((5-0)^(2)+a^(2))` <br> `=(-q)/(4piin_(0)sqrt(25^(2)+a^(2))+q/(4pi in_(0)sqrt(25+a^(2))` <br> =0 <br> Electrostatic potentials, `V_(2)`, at point `(-7,0,0)` is given by <br> V_(2)=(-q)/(4pi in_(0))1/(sqrt(-7)^(2)+(-a)^(2))+q/(4pi in_(0))1/(sqrt((-7)^(2)+(a)^(2)))` <br> `=(-q)/(4pi in_(0))sqrt(49+a^(2))+q/(4pi in_(0))1/(sqrt(49+a^(2)))` <br> =0 <br> Hence, no work is done in moving a small test charge from point (5,0,0) to point (-7,0,0) along the x-axis . The answer does not change because work doen by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.

Related Video

18.5 K+ Views | 900+ Likes

24.6 K+ Views | 15.5 K+ Likes

3:34

57.1 K+ Views | 2.8 K+ Likes

3:54

99.4 K+ Views | 61.5 K+ Likes

3:28

48.5 K+ Views | 2.4 K+ Likes

8:36

351.9 K+ Views | 17.5 K+ Likes

8:41

57.2 K+ Views | 2.8 K+ Likes

bottom open in app