Explanation
The electric field , $$\displaystyle E=\dfrac{1}{4\pi\epsilon_0} \dfrac{q}{r^2}=9\times 10^9\times \dfrac{2\times 10^{-6}}{(2-1)^2+(3-2)^2+(4-3)^2}=6000 NC^{-1}$$
and potential, $$\displaystyle V=\dfrac{1}{4\pi\epsilon_0} \dfrac{q}{r}=9\times 10^9\times \dfrac{2\times 10^{-6}}{\sqrt{(2-1)^2+(3-2)^2+(4-3)^2}}=6000 \sqrt 3 NC^{-1}$$
Initial potential energy: $$\dfrac { 2k{ q }_{ 1 }{ q }_{ 2 } }{ r } =\quad (2\times 9\times { 10 }^{ 9 }\times \dfrac { 1 }{ 3 } \times { 10 }^{ -6 }\times \dfrac { -1 }{ 3 } \times { 10 }^{ -6 })/(1)=-2\times { 10 }^{ -3 }$$
Initial kinetic energy is 10^(-3) J.
Let the body is at r distance from both the charges (will be at equal distance from both as both charges are same)
Final Kinetic energy is zero (as body stopped).
By equating Initial Total Energy & Final Total Energy:
$$-2\times { 10 }^{ -3 }+{ 10 }^{ -3 }=\quad \dfrac { 2k{ q }_{ 1 }{ q }_{ 2 } }{ r } =(2\times 9\times { 10 }^{ 9 }\times \dfrac { 1 }{ 3 } \times { 10 }^{ -6 }\times \dfrac { -1 }{ 3 } \times { 10 }^{ -6 })/(r)\\ -1\times { 10 }^{ -3 }=-2\times { 10 }^{ -3 }/(r)\\ \therefore \quad r=2$$
Therefore distance traveled:
$$\sqrt { { ({ r }_{ f }^{ 2 }- }{ r }_{ i }^{ 2 }) } =\sqrt { 4-1 } =\sqrt { 3 } $$
There is no potential gradient along any direction parallel to the surface, and no electric field is parallel with the surface, This means electric field are always at right angle to the equipotential surface.
Please disable the adBlock and continue. Thank you.