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. $\frac { 2 \mathrm { qd } } { 4 \pi \varepsilon _ { 0 } \left( \mathrm {~d} ^ { 2 } + \mathrm { a } ^ { 2 } \right) ^ { 3 / 2 } }$ $\widehat{k}$
• B. $\frac { \mathrm { qd } } { 4 \pi \varepsilon _ { 0 } \left( \mathrm {~d} ^ { 2 } + \mathrm { a } ^ { 2 } \right) ^ { 3 / 2 } }$ $\widehat{k}$
• C. $\frac { - 2 \mathrm { qd } } { 4 \pi \varepsilon _ { 0 } \left( \mathrm {~d} ^ { 2 } + \mathrm { a } ^ { 2 } \right) ^ { 3 / 2 } }$ $\widehat{k}$
• D. $\frac { - \mathrm { qd } } { 4 \pi \varepsilon _ { 0 } \left( \mathrm {~d} ^ { 2 } + \mathrm { a } ^ { 2 } \right) ^ { 3 / 2 } }$ $\widehat{k}$

Answer 1

Answer: (C)

\frac { - 2 \mathrm { qd } } { 4 \pi \varepsilon _ { 0 } \left( \mathrm {~d} ^ { 2 } + \mathrm { a } ^ { 2 } \right) ^ { 3 / 2 } }$$\widehat{k}

Answer 2

Use formula E =

\ E = $\frac { 1 } { 4 \pi \varepsilon _ { 0 } }$ $\frac { \mathrm { q } \times 2 \mathrm {~d} } { \left( \mathrm { a } ^ { 2 } + \mathrm { d } ^ { 2 } \right) ^ { 3 / 2 } }$

\ $\overrightarrow { \mathrm { E } }$ = $\frac { 1 } { 4 \pi \varepsilon _ { 0 } }$ (–)