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Chapter 22: Q. 31 - Excercises And Problems (page 625)

The electric field 2.0 cm from a small object points away from the object with a strength of 270,000 N/C. What is the object’s charge?

Short Answer

Expert verified

Q=12.0×10−9C

Step by step solution

01

Given information

It is given that,

Electric field strength,E=270000N/C

Distance from a small object,r=2.0cm=0.02m

02

Explanation

Electric field at a point is given by : E=kQr2

Q is the charge on an object

localid="1650565043933" Q=Er2kQ=270000N/C×(0.02m)29×109Nm2/C2Q=12.0×10−9C

So, the charge on the object is localid="1650565055607" 12.0×10−9C. Hence, this is the required solution.

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