Animated Solution for Physics - Electrostatics: The electric field in a region is given by $\mathbf{E}=(Ax+B)\hat{\mathbf{i}}$, where $E$ is in ${\text{NC}}^{-1}$ and $x$ is in metres. The values of constants are $A=20$ SI unit and $B=10$ SI unit. If the potential at $x=1$ is $V_1$ and that at $x=-5$ is $V_2$, then $V_1-V_2$ is

In electrostatics, the electric field and electric potential are intimately connected. The electric field is essentially the negative gradient of the electric potential. This means that if we know the electric field in a region, we can find the potential difference between any two points by integrating the field along the path connecting them.

The fundamental equation governing this relationship is $dV=-\mathbf{E}\cdot d\mathbf{x}$. The negative sign is crucial here; it physically signifies that the electric field always points in the direction of decreasing electric potential. Imagine a ball rolling down a hill—the field points "downhill" towards lower potential.

To find the potential difference between two specific points, we integrate our differential equation. Integrating from an initial position $x_1$ to a final position $x_2$ gives us $V_2-V_1=-{\int }_{x_1}^{x_2}\mathbf{E}\cdot d\mathbf{x}$.

However, the question specifically asks for $V_1-V_2$. By multiplying both sides of our integrated equation by $-1$, we elegantly absorb the negative sign on the right side. This leaves us with a much cleaner expression: $V_1-V_2={\int }_{x_1}^{x_2}\mathbf{E}\cdot d\mathbf{x}$.

Now, we substitute the given electric field $\mathbf{E}=(20x+10)\hat{\mathbf{i}}$ into our integral. Since the displacement $d\mathbf{x}$ is also along the x-axis ($dx\hat{\mathbf{i}}$), the dot product simplifies to just the product of their magnitudes. Our integral becomes $V_1-V_2={\int }_1^{-5}(20x+10)dx$.

Performing the integration is straightforward polynomial calculus. The integral of $20x$ is $10x^2$, and the integral of $10$ is $10x$. Factoring out the $10$ for simplicity, we get $10[x^2+x]$ evaluated from $1$ to $-5$.

Finally, we carefully substitute our upper and lower limits. Plugging in $-5$ gives $((-5{)}^2-5)=20$, and plugging in $1$ gives $(1^2+1)=2$. Subtracting the lower limit evaluation from the upper limit evaluation yields $20-2=18$. Multiplying this by our factored $10$ gives the final potential difference of $180\text{ V}$.