Question

If the electric flux entering and leaving an enclosed surface respectively is $ {{\phi }_{1}} $ and $ {{\phi }_{2}} $ the electric charge inside the surface will be

• A. $ ({{\phi }_{2}}-{{\phi }_{1}}){{\varepsilon }_{0}} $
• B. $ ({{\phi }_{1}}+{{\phi }_{2}})/{{\varepsilon }_{0}} $
• C. $ ({{\phi }_{2}}-{{\phi }_{1}})/{{\varepsilon }_{0}} $
• D. $ ({{\phi }_{1}}+{{\phi }_{2}}){{\varepsilon }_{0}} $

Answer 1

Answer: (A)

[From Gauss law, $ \frac{\text{Charge}\,\,\text{enclosed}}{{{}_{\text{0}}}} $ = Flux leaving the surface] $ \frac{q}{{{\varepsilon }_{0}}}={{\phi }_{2}}-{{\phi }_{1}} $ $ \Rightarrow $ $ q=({{\phi }_{2}}-{{\phi }_{1}}){{\varepsilon }_{0}} $