Problem 34 - A Complete Resource Book in Physics for JEE Main

Question

The electric field at the origin is along the positive \(X\)-axis. A small circle is drawn with the centre at the origin cutting the axes at points \(A, B, C\), and \(D\) having coordinates \((a, 0)(0, a),(-a, 0),(0,-a)\), respectively. Out of the points on the periphery of the circle, the potential is minimum at (A) \(A\) (B) \(B\) (C) \(C\) (D) \(D\)

Step-by-Step Solution

Answer

The point with minimum potential is (C) C, as it is opposite to the electric field direction along the positive x-axis, and all distances from the origin are equal.

Step 1: Electric potential formula

The electric potential V at a point due to the electric field can be found using the formula: \(V = E \cdot d\) Where E is the electric field and d is the distance between the point and the origin. Since we know the electric field is along the positive x-axis, we only need to find the distance d to each point and compare them.

Step 2: Calculate distance d for each point

We'll use the distance formula to find the distance d from the origin to each of the given points A, B, C, and D. \(Distance (d) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) For point A, the distance is: \(d_A = \sqrt{(a - 0)^2 + (0 - 0)^2} = a\) For point B, the distance is: \(d_B = \sqrt{(0 - 0)^2 + (a - 0)^2} = a\) For point C, the distance is: \(d_C = \sqrt{(-a - 0)^2 + (0 - 0)^2} = a\) For point D, the distance is: \(d_D = \sqrt{(0 - 0)^2 + (-a - 0)^2} = a\) Notice that all distances are equal to 'a'.

Step 3: Compare the electric potentials at each point

Since all distances are equal, the electric potentials at points A, B, C, and D will be the same: \(V_A = E \cdot a\) \(V_B = E \cdot a\) \(V_C = E \cdot a\) \(V_D = E \cdot a\) However, we are asked to find the minimum potential, and we know that the electric field direction is along the positive x-axis. Thus, minimum potential should occur at the point opposite to the electric field direction. Therefore, the point with minimum potential is: (C) C

Key Concepts

Electric Field

The electric field is a vector quantity that permeates the space around electric charges. It represents the force that a charge would experience per unit of charge. Symbolically, we express it as \(\textbf{E}\). The direction of the electric field at any point is the direction of the force that would be exerted on a positive test charge placed at that point. In simple terms, it points away from positive charges and toward negative charges.

When visualizing an electric field, one could imagine invisible lines radiating from a charge, much like the rays from the sun. These lines extend from positive to negative, forming what we call electric field lines. At any specific location, if we draw a tangent to these lines, it gives us the direction of the field at that point. The strength or intensity of the electric field is greatest closer to the charge and diminishes with distance.

When establishing the connection between electric fields and potential at various points, as in the textbook exercise, the understanding of electric field lines provides a logical path to visualizing why potential might differ at various points in a field, given their distances and positions relative to the source of the field.

Electric Potential Formula

Electric potential, often denoted by \(V\), is the potential energy per unit charge associated with a static (time-invariant) electric field. It provides a scalar measure of how much work an electric field would do in moving a charge from one point to another. The formula for electric potential due to a point charge at a distance \(r\) is given by \(V = \frac{kQ}{r}\), where \(k\) is Coulomb's constant, \(Q\) is the charge, and \(r\) is the distance from the charge.

However, the context within the given exercise deals with a uniform electric field, and in such a scenario, we use the formula \(V = E \cdot d\), where \(E\) represents the electric field strength and \(d\) is the distance along the direction of the field from the reference point (in this case, the origin). This formula implies that potential difference arises due to the work done against the electric field to move a charge.

In situations where distances are equal, as illustrated in the problem-solving steps, the electric potential would be identical at all points if only magnitude is considered. Yet, the exercise also involves the direction of the electric field, which plays a critical role in determining the location of minimum potential.

JEE Main Physics

The Joint Entrance Examination (JEE) Main is a prominent engineering entrance test in India, where Physics is a key subject in its syllabus. The concepts of Electric Field and Electric Potential are integral to this exam's physics section.

Problems similar to the textbook exercise not only test a student's ability to apply physics formulas but also their understanding of foundational concepts, such as the relationship between force, field, and potential. Such questions require both conceptual clarity and problem-solving skills, which are critical in JEE Main Physics.

For students preparing for this exam, mastery over electric phenomena is necessary. This includes not only knowing and manipulating equations but also deciphering physical situations, visualizing electric fields, and comprehending how charges interact within these fields. Moreover, practice with diverse problems, such as the one illustrated, helps students hone their analytical thinking and improves their readiness for the JEE Main Physics challenges they will face.