The electric potential (V) in volts varies with x (in metre) according to the relation V=5+4x2. The force experienced by a negative charge of 2 μC located at x=0.5m is

V=5+4x2,E=−dvdx=−8x=−8×(0.5)=−4v∴F=Eq=(−4)(−2)=8×10−6N

The electric potential (V) in volts varies with x (in metre) according to the relation V=5+4x2. The force experienced by a negative charge of 2 μC located at x=0.5m is

V=5+4x2,E=−dvdx=−8x=−8×(0.5)=−4v∴F=Eq=(−4)(−2)=8×10−6N

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The electric potential (V) in volts varies with x (in metre) according to the relation V=5+4x². The force experienced by a negative charge of 2 $μ$ C located at x=0.5m is

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2 x 10^-6 N

4 x 10^-6 N

8 x 10^-6 N

6 x 10^-6 N

answer is D.

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$\begin{array}{l} V = 5 + 4 x^{2}, E = \frac{- dv}{dx} = - 8 x = - 8 \times (0.5) = - 4 v \\ ∴ F = Eq = (- 4) (- 2) = 8 \times 10^{- 6} N \end{array}$

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The electric potential (V) in volts varies with x (in metre) according to the relation V=5+4x2. The force experienced by a negative charge of 2 μC located at x=0.5m is

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The electric potential (V) in volts varies with x (in metre) according to the relation V=5+4x². The force experienced by a negative charge of 2 $μ$ C located at x=0.5m is