Question 31: Electrostats & Capacitance

Question

Electric field in a region is given by $\vec{E} = Ax\hat{i} + By\hat{j}$, where $A = 10	ext{ V/m}^2$ and $B = 5	ext{ V/m}^2$. If the electric potential at a point $(10, 20)$ is $500	ext{ V}$, then the electric potential at origin is ______ $	ext{V}$.

Accepted Answer

The total work done in moving a charge between two points in an electric field is independent of the path taken and depends only on the potential difference between those points.Problem BreakdownGiven Electric Field: $\vec{E} = Ax\hat{i} + By\hat{j}$Constants: $A = 10 	ext{ V/m}^2$, $B = 5 	ext{ V/m}^2$Potential at Point (10, 20): $V = 500 	ext{ V}$Goal: Find the potential at the origin $(0, 0)$, denoted as $V_0$.Step-by-Step Calculation1. Relation between Field and Potential:The change in electric potential $dV$ is related to the electric field $\vec{E}$ $by the equation:$$dV = -\vec{E} \cdot d\vec{r} = -(E_x dx + E_y dy)$$2. Substituting the Components:$$dV = -(Ax dx + By dy)$$3. Integration:Integrate from the origin $(0, 0)$ where the potential is $V_0$, to the point $(10, 20)$ where the potential is $500 	ext{ V}$:$$\int_{V_0}^{500} dV = -\int_{0}^{10} Ax \, dx - \int_{0}^{20} By \, dy$$$$[V]_{V_0}^{500} = -A \left[ \frac{x^2}{2} ight]_0^{10} - B \left[ \frac{y^2}{2} ight]_0^{20}$$4. Solving for $V_0$:$$500 - V_0 = -10 \left( \frac{100}{2} ight) - 5 \left( \frac{400}{2} ight)$$$$500 - V_0 = -500 - 1000$$ $$500 - V_0 = -1500$$ $$V_0 = 500 + 1500$$ $$V_0 = 2000 	ext{ V} $$\$$Conclusion:The electric potential at the origin is 2000 V.$$

Answer

A. $1000$

Answer

B. $0$

Answer

C. $2000$

Answer

D. $500$

Electrostats & Capacitance

Electric field in a region is given by \vec{E} = Ax\hat{i} + By\hat{j}, where A = 10\text{ V/m}^2 and B = 5\text{ V/m}^2. If the electric potential at a point (10, 20) is 500\text{ V}, then the electric potential at origin is ______ \text{V}.

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