Question

Two point charges q and –q are at positions (0, 0, d) and (0, 0, –d) respectively. What is the electric field at (a, 0, 0) ?

• A. $\frac{2qd}{4\pi\varepsilon_{0}(d^{2} + a^{2})^{3/2}}\widehat{k}$
• B. $\frac{qd}{4\pi\varepsilon_{0}(d^{2} + a^{2})^{3/2}}\widehat{k}$
• C. $\frac{- 2qd}{4\pi\varepsilon_{0}(d^{2} + a^{2})^{3/2}}\widehat{k}$
• D. $\frac{- qd}{4\pi\varepsilon_{0}(d^{2} + a^{2})^{3/2}}\widehat{k}$

Answer 1

Answer: (C)

$\frac{- 2qd}{4\pi\varepsilon_{0}(d^{2} + a^{2})^{3/2}}\widehat{k}$

Answer 2

r = (a² + d²)^1/2 at angle 180 – 2 q

E_net = 2E_A cos $\frac { 1 } { 2 }$ (180 – 2q) = 2E_Acos(90 – q)

= 2E_A sinq = $\frac { 2 \mathrm { kq } } { \mathrm { r } ^ { 2 } }$ $\frac { \mathrm { d } } { ( \mathrm { r } ) }$

= $\frac { 2 \mathrm { kqd } } { \left( \mathrm { a } ^ { 2 } + \mathrm { d } ^ { 2 } \right) ^ { 3 / 2 } }$ in –z direction.