Question

Two point charges q1=+ 3.2 nC and q2= -5.3 nC are 0.2 m apart. Point A is midway between them; point B is 0.09 m from q1 and 0.07 from q2. Take the electric potential to be zero at infinity. Find (a) the potential at point A (b) the potential at point B (c) the work done by the electric field on a charge of 3.5 nC that travels from point A to point B.

Answer 1

(a) The potential of a charge can be calculated as $\varphi = k\cdot\dfrac{q}{r}.$ The total potential in A is $k\cdot\dfrac{q_1}{r_1} + k\cdot\dfrac{q_2}{r_2} = 9\cdot10^9\,\mathrm{V\cdot m/C}\cdot\left(\dfrac{3.2\cdot10^{-9}\,\mathrm{C}}{0.1\,\mathrm{m}} + \dfrac{-5.3\cdot10^{-9}\,\mathrm{C}}{0.1\,\mathrm{m}} \right) = -189\,\mathrm{V}.$the

(b) We can see that there can be such a point. The sum of distance from q₁ to B and from q₂ to B cannot be less than the distance between q₁ and q₂. The potential in such a virtual point B can be calculated similarly:

$k\cdot\dfrac{q_1}{r_1} + k\cdot\dfrac{q_2}{r_2} = 9\cdot10^9\,\mathrm{V\cdot m/C}\cdot\left(\dfrac{3.2\cdot10^{-9}\,\mathrm{C}}{0.09\,\mathrm{m}} + \dfrac{-5.3\cdot10^{-9}\,\mathrm{C}}{0.07\,\mathrm{m}} \right) = -361\,\mathrm{V}.$

(c) The work can be calculated as

$W_{A\to B} = q\cdot(\varphi_A-\varphi_B) = 3.5\cdot10^{-9}\,\mathrm{C}\cdot (-189\,\mathrm{B}-(-361\,\mathrm{V})) = 6\cdot10^{-7}\,\mathrm{J}.$