JEE MAIN - Physics (2014 (Offline) - No. 16)
Assume that an electric field $$\overrightarrow E = 30{x^2}\widehat i$$ exists in space. Then the potential difference $${V_A} - {V_O},$$ where $${V_O}$$ is the potential at the origin and $${V_A}$$ the potential at $$x=2$$ $$m$$ is :
$$120$$ $$J/C$$
$$-120$$ $$J/C$$
$$-80$$ $$J/C$$
$$80$$ $$J/C$$
Explanation
Potential difference between any two points in an electric field is given by,
$$dV = - \overrightarrow E .\overrightarrow {dx} $$
$$\int\limits_{{V_O}}^{{V_A}} {dV = - \int\limits_0^2 {30{x^2}} } dx$$
$${V_A} - {V_O} = \left[ {10{x^3}} \right]_{0}^2 = - 80J/C$$
$$dV = - \overrightarrow E .\overrightarrow {dx} $$
$$\int\limits_{{V_O}}^{{V_A}} {dV = - \int\limits_0^2 {30{x^2}} } dx$$
$${V_A} - {V_O} = \left[ {10{x^3}} \right]_{0}^2 = - 80J/C$$
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