Chapter 15: Q2Q (page 616)

A rod is $2.5 m$ long. Its charge is $- 2 \times 10^{- 7} C$ . The observation location is $4 cm$ from the rod, in the mid plane. In the expression
$E = \frac{1}{4 π ε_{0}} \frac{Q}{r \sqrt{r^{2} + {(\frac{L}{2})}^{2}}}$
what is $r$ in meters?

Short Answer

Expert verified

The value of $r$ is $0.04 m .$

Step by step solution

Given data

A uniformly charged rod of length $L$ .

Distance from the rod of the point where the electric field is observed is $4 c m$ .

Step 2:Electric field due to a uniformly charged rod

The electric field due to a rod of length $L$ and charge $Q$ uniformly distributed at a distance $r$ from it on its mid plane is,

role="math" localid="1668497952679" $E = \frac{1}{4 π ε_{0}} \frac{Q}{r \sqrt{r^{2} + {(\frac{L}{2})}^{2}}}$

Step 3:Determining the distance r

From equation (I), it is seen that $r$ is the distance of the observation point from the rod, that is,

$r = 4 cm = 4 \cdot (1 cm \times \frac{1 m}{100 cm}) = 0.04 m$

Thus, $r$ is meters is $0.04 m$ .

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A rod is $2.5 m$ long. Its charge is $- 2 \times 10^{- 7} C$ . The observation location is $4 cm$ from the rod, in the mid plane. In the expression
$E = \frac{1}{4 π ε_{0}} \frac{Q}{r \sqrt{r^{2} + {(\frac{L}{2})}^{2}}}$
what is $r$ in meters?

Short Answer

Step by step solution

Given data

Step 2:Electric field due to a uniformly charged rod

Step 3:Determining the distance r

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A rod is 2.5m long. Its charge is -2×10-7C. The observation location is 4cm from the rod, in the mid plane. In the expressionE=14πε0Qrr2+(L2)2what isr in meters?

Short Answer

Step by step solution

Given data

Step 2:Electric field due to a uniformly charged rod

Step 3:Determining the distance r

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

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Study anywhere. Anytime. Across all devices.

A rod is $2.5 m$ long. Its charge is $- 2 \times 10^{- 7} C$ . The observation location is $4 cm$ from the rod, in the mid plane. In the expression
$E = \frac{1}{4 π ε_{0}} \frac{Q}{r \sqrt{r^{2} + {(\frac{L}{2})}^{2}}}$
what is $r$ in meters?