Problem 58 - Master Resource Book in JEE Main Physics

Question

A uniform electric field of \(100 \mathrm{~N} / \mathrm{C}\) exist in the vertically downward direction. The increase in the electric potential as one goes up the through a height of \(50 \mathrm{~cm}\) is (a) \(10 \mathrm{~V}\) (b) \(5 \mathrm{~V}\) (c) \(0 \mathrm{~V}\) (d) \(0.5 \mathrm{~V}\)

Step-by-Step Solution

Answer

The calculated increase in electric potential is 50 V.

Step 1: Understand the Problem

We need to determine the change in electric potential as we move up 50 cm against a uniform electric field. The field strength is given as 100 N/C, acting in the downward direction.

Step 2: Formula for Change in Electric Potential

The change in electric potential, 90V, in a uniform electric field is given by the formula \( \Delta V = -E \, \Delta h \), where \( E \) is the magnitude of the electric field and \( \Delta h \) is the change in height along the direction of the field. Here, since the field is downward, moving upward means \( \Delta h \) is positive.

Step 3: Plug in the Values

Substitute the given values into the formula: the magnitude of the electric field \( E = 100 \, \text{N/C} \) and the change in height \( \Delta h = 50 \, \text{cm} = 0.5 \, \text{m} \). Thus, \( \Delta V = -100 \, \text{N/C} \times 0.5 \, \text{m} \).

Step 4: Calculate the Change in Electric Potential

Calculate \( \Delta V = -100 \, \times 0.5 = -50 \, \text{V} \). Since we are moving against the direction of the field, the electric potential increases by 50 V.

Step 5: Determine the Correct Option

The calculated change in electric potential is 50 V. Re-evaluate the given options to find that none exactly match our calculated answer, indicating a re-evaluation mistake. Check compatibility with provided options or initial statements.

Key Concepts

Uniform Electric Field

Imagine an area where the electric field strength is the same at every point; this is called a uniform electric field. In simpler terms, a uniform electric field has parallel lines with equal spacing illustrating constant force on a positive test charge throughout the field.

• In a uniform electric field, the direction of the field lines indicates where a positive charge would move if placed in the field.
• The field lines are usually straight and equally spaced in a uniform field.
• In our exercise, the electric field is directed downward, which means that a positive charge would naturally move down unless acted upon by another force.

Understanding uniform electric fields helps us grasp how forces operate consistently across a particular region, creating predictable effects on charges within that field.

Electric Field Strength

Electric field strength is a way of quantifying the influence an electric field exerts on a charge at a point in space. It is denoted by the letter "E" and measured in Newtons per Coulomb (N/C).

• It defines how much force would be exerted on a positive charge placed in the field.
• The stronger the electric field, the greater the force experienced by the charge.
• In our example, the uniform electric field strength is given as 100 N/C.

Understanding the electric field strength helps in predicting the behavior of charges within the field. It provides insight into how much work is required to move a charge against the field, as indicated by the changes in electric potential.

Change in Height

Change in height in the context of electric potential refers to moving a charge vertically within an electric field.

• This movement affects the electric potential energy of the charge.
• When you move upward in a downward uniform electric field, like in our exercise, you're moving against the field's direction.
• This results in an increase in electric potential energy. This is because doing work against the electric field increases potential energy.

Given the uniform field strength of 100 N/C, and a change in height of 0.5 meters, our problem demonstrates how these factors result in a calculated change in electric potential of -50 V, indicating an increase since it's against the field direction.