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(A) The potential difference $V_A - V_B$ is given by the area under the $E-x$ graph between the points $x_A$ and $x_B$.$V_2 - V_6 = \int_{2}^{6} E \,

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) The potential difference $V_A - V_B$ is given by the area under the $E-x$ graph between the points $x_A$ and $x_B$.$V_2 - V_6 = \int_{2}^{6} E \, dx$This integral represents the area under the $E-x$ graph from $x = 2 \, m$ to $x = 6 \, m$.The area consists of a rectangle from $x = 2$ to $x = 4$ and a triangle from $x = 4$ to $x = 6$.Area of rectangle = $base 	imes height = (4 - 2) 	imes 10 = 2 	imes 10 = 20 \, V$.Area of triangle = $\frac{1}{2} 	imes base 	imes height = \frac{1}{2} 	imes (6 - 4) 	imes 10 = \frac{1}{2} 	imes 2 	imes 10 = 10 \, V$.Total potential difference = $20 + 10 = 30 \, V$.

The figure shows the variation of electric field intensity $E$ versus distance $x$. What is the potential difference between the points at $x = 2 \, m$ and $x = 6 \, m$ from $O$ (in $V$)?

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