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Chapter 22: Problem 23

An electric field is given by \(\vec{E}=E_{0} \hat{\jmath},\) where \(E_{0}\) is a constant. Find the potential as a function of position, taking \(V=0\) at \(y=0\)

Short Answer

Expert verified
The electric potential as a function of position is \(V(y) = -E_{0}y\).

Step by step solution

01

Formulate the equation

The relationship between the electric field \(\vec{E}\) and electric potential \(V\) is given by \(\vec{E} = -\nabla V\). In terms of cartesian coordinates, this becomes \(\vec{E} = -\frac{dV}{dx}\hat{\imath} -\frac{dV}{dy}\hat{\jmath} -\frac{dV}{dz}\hat{k}\). In this case, the electric field has only the y-component, \(E_{0} \hat{\jmath}\). So, \(-\frac{dV}{dy}\hat{\jmath} = E_{0} \hat{\jmath}\).
02

Solve the differential equation

Rearrange the equation \( -\frac{dV}{dy}\hat{\jmath} = E_{0}\hat{\jmath}\) to get \(\frac{dV}{dy} = -E_{0}\). Now, integrate both sides with respect to \(y\) to find \(V(y)\), the potential as a function of position.
03

Apply boundary condition

The problem states that at \(y=0\), \(V=0\). This will act as the boundary condition to find the constant of integration. Since \(V(0) = 0\), the constant of integration is zero.
04

Find Potential as function of position

With the constant of integration being zero, the potential function becomes \(V(y) = -E_{0}y\). So, the electric potential \(V\) as a function of position \(y\) is \(V(y) = -E_{0}y\).

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