Following figure indicates the electric potential V as a function through 4 regions on x-axis. Which of the following is correct for the electric field E in these regions?
Show Hint
Field is minus the slope of the V-x graph. Flat parts give zero field, and the steeper part gives the larger field.
Step 1: Understand the figure
The graph shows potential \(V\) against position \(x\) split into four regions. In region 1 the line is flat. In region 2 the potential rises steeply to point C. In region 3 it is flat again (C to D). In region 4 it falls steeply from D down to zero at E.
Step 2: Use the field-potential relation
\(E=-\frac{dV}{dx}\). So the size of the field in a region is the size of the slope of the V-x graph there.
Step 3: Regions 1 and 3
Both are horizontal lines. The slope is zero, so \(E_1=0\) and \(E_3=0\). Hence \(E_1=E_3\).
Step 4: Regions 2 and 4
Region 2 rises by \(V_C-V_B\) over a width \(\Delta x_2\). Region 4 falls by the full \(V_C\) over a width \(\Delta x_4\). From the figure region 4 is narrower and has the larger potential change, so its slope is steeper. This gives \(E_2<E_4\).
Step 5: Check the options
(A) says all four fields increase in order, but \(E_1=E_3=0\) cannot be strictly increasing. (B) says \(E_2=E_4\), which is false. (D) says the fields decrease in order, which is also false. Only (C) fits.
Final Answer:
\(E_1=E_3\) and \(E_2<E_4\), option (C).
\[ \boxed{E_1=E_3\ \text{and}\ E_2<E_4} \]
