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Fundamentals Of Physics
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Short Answer

In a Millikan oil-drop experiment (Module 22-6), a uniform electric field of 1.92 × 105 N/C is maintained in the region between two plates separated by 1.50 cm . Find the potential difference between the plates.

The potential difference between the plates is, 2.9 kV.

See the step by step solution

Step by Step Solution

Step 1: Identification of the given data

The uniform electric field is, E=1.92 × 105 N/C.

The separation between the two plates is, R=0.015 m

Step 2: Understanding the concept

According to the definition, the electric potential gradient ‘E’ is given as:

role="math" localid="1662732704306" E=νR

Or

V=E×RV=ER

Step 3: Calculate the potential difference between the plates

The potential difference between the plates is expressed is,

V=ER

Substitute all the value in the above equation.

V=1.92 × 105 N/C × 0.015 m=2.88×103 V=2.9 kV

Hence the potential difference between the plates is, 2.9 kV .

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