Problem 100
Question
A uniform electric field \(E_{0}\) exists in a region at angle \(45^{\circ}\) with
\(x\)-axis. There are two point \(A(a, 0)\) and \(B(0, b)\) having potential \(V_{A}\)
and \(V_{B}\), respectively, then
(A) \(V_{A}>V_{B}\) if \(a>b\)
(B) \(V_{A}=V_{B}\) if \(a=b\)
(C) \(V_{A}>V_{B}\) if \(a
Step-by-Step Solution
Verified Answer
(B) \(V_A = V_B\) if \(a=b\)
(D) \(V_A < V_B\) if \(a>b\) and electric field \(E_0\) is positive.
(A) \(V_A > V_B\) if \(a>b\) and electric field \(E_0\) is negative.
(C) \(V_A > V_B\) if \(a
1Step 1: Calculate electric field components
The electric field is at a \(\displaystyle 45^\circ\) angle with the x-axis. We can resolve it into two components, i.e., x-component (\(\displaystyle E_x\)) and y-component (\(\displaystyle E_y\)).
Using trigonometry;
\[
E_x = E_0 \cos 45^\circ = \frac{E_0}{\sqrt{2}}
\]
\[
E_y = E_0 \sin 45^\circ = \frac{E_0}{\sqrt{2}}
\]
2Step 2: Calculate Potential difference between points A and B
The potential difference between the two points can be calculated using the expression \(\displaystyle \Delta V = -\int_a^b \vec{E} · \vec{dl}\), where the dot product represents the work done by the electric field against the displacement between the two points.
In this case, from point A to B, we need to move along the x-direction for a distance 'a' and then y-direction for distance 'b'.
So, potential difference produced due to displacement A to B;
\[
\Delta V = - \int_0^a E_x dx - \int_0^b E_y dy
\]
Now, substitute the values of \(\displaystyle E_x\) and \(\displaystyle E_y\):
\[
\Delta V = -\int_0^a \frac{E_0}{\sqrt{2}} dx - \int_0^b \frac{E_0}{\sqrt{2}} dy
\]
3Step 3: Evaluate the integrals and simplify the expression:
Evaluating the integrals and simplifying the expression:
\[
\Delta V = -\frac{E_0}{\sqrt{2}} \left[a + b\right]
\]
By definition, we have that \(\displaystyle V_A - V_B = \Delta V\).
Thus,
\[
V_A - V_B = -\frac{E_0}{\sqrt{2}}\left[a+b\right]
\]
4Step 4: Compare the potentials and draw conclusions
(A) When \(a > b\), we have \(V_A > V_B\) if \(\displaystyle \frac{E_0}{\sqrt{2}}(a+b)\) is negative. This is only true if the electric field \(\displaystyle E_0\) is negative.
(B) When \(a = b\), we have \(V_A = V_B\) as \(\displaystyle \frac{E_0}{\sqrt{2}}(a+a) = 0\)
(C) When \(a < b\), we have \(V_A > V_B\) if \(\displaystyle \frac{E_0}{\sqrt{2}}(a+b)\) is negative. This is only true if the electric field \(\displaystyle E_0\) is negative.
(D) When \(a > b\), we have \(V_A < V_B\) if \(\displaystyle \frac{E_0}{\sqrt{2}}(a+b)\) is positive. This is only true if the electric field \(\displaystyle E_0\) is positive.
In conclusion, the relation between the potentials depends on the direction and magnitude of the electric field. So, we can't determine the correct relation without knowing the nature of the electric field.
Other exercises in this chapter
Problem 97
Capacitance of a capacitor becomes \(\frac{4}{3}\) times its original value if a dielectric slab of thickness \(t=d / 2\) is inserted between the plates \((d=\)
View solution Problem 99
Three point charges \(q,-2 q\), and \(-2 q\) are placed at the vertices of an equilateral triangle of side \(a\). The work done by some external force to slowly
View solution Problem 101
A non-conducting solid sphere of radius \(R\) is uniformly charged. The magnitude of the electric field due to the sphere at a distance \(r\) form its centre (A
View solution Problem 102
A point charge \(q\) is placed at origin. Let \(\vec{E}_{A}, \vec{E}_{B}\) and \(\vec{E}_{C}\) be the electric field at three points \(A(1,2,3)\), \(B(1,1,-1)\)
View solution