• Point charge (Q) = 10 μC = 10 × 10⁻⁶ C
• Potential (V) = 1000 V
• Distance (r) = ?
[tex]\implies V = \dfrac{KQ}{r} \\ [/tex]
[tex]\implies r= \dfrac{KQ}{V}[/tex]
[tex]\implies r= \dfrac{9 \times {10}^{9} \times 10 \times {10}^{ - 6} }{1000} \\ [/tex]
[tex]\implies r= \dfrac{90 \times {10}^{3} }{1000} \\ [/tex]
[tex]\implies r= \dfrac{90 \times {10}^{3} }{ {10}^{3} } \\ [/tex]
[tex]\implies r= 90 \times {10}^{3} \times {10}^{ - 3} \\ [/tex]
[tex]\implies\bf r= 90\:m \\ [/tex]
Hence,the option B) 90 m is the correct answer.
