A uniform electric field of 200 Vm-1 is directed at 60- with the positive x-axis as shown in figure- If OA - 2 m- the potential difference V0 - VA is

The correct option is A 200 VIn the given figure : OB = OAcos60^0 = 2 ×1/2 = 1 m V_B V_O = e( OB) = 200 × 1 = 200V V_B V_O = 200V [ V_ ...

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Byju's Answer

Standard XII

Physics

Potential Energy - External Viewpoint

A uniform ele...

Question

A uniform electric field of $200 V m^{- 1}$ is directed at $60^{\circ}$ with the positive x-axis as shown in figure. If OA = 2 m, the potential difference $V_{0} - V_{A}$ is

- 200 V

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50 V

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100 V

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- 100 V

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Solution

The correct option is A - 200 V
In the given figure $: -$
$O B = O A cos 60^{0} = 2 \times \frac{1}{2} = 1 m$
$V_{B} - V_{O} = e (O B) = 200 \times 1 = 200 V$
$V_{B} - V_{O} = 200 V [V_{B} - V_{A}]$
$\Rightarrow V_{O} - V_{A} = - 200 V$
Hence,
option $(A)$ is correct answer.

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A uniform electric field of 200Vm−1 is directed at 60∘ with the positive x-axis as shown in figure. If OA = 2 m, the potential difference V0−VA is

The correct option is A - 200 VIn the given figure:−OB=OAcos600=2×12=1mVB−VO=e(OB)=200×1=200VVB−VO=200V[VB−VA]⇒VO−VA=−200VHence,option (A)is correct answer.

A uniform electric field of $200 V m^{- 1}$ is directed at $60^{\circ}$ with the positive x-axis as shown in figure. If OA = 2 m, the potential difference $V_{0} - V_{A}$ is

The correct option is A - 200 V
In the given figure $: -$
$O B = O A cos 60^{0} = 2 \times \frac{1}{2} = 1 m$
$V_{B} - V_{O} = e (O B) = 200 \times 1 = 200 V$
$V_{B} - V_{O} = 200 V [V_{B} - V_{A}]$
$\Rightarrow V_{O} - V_{A} = - 200 V$
Hence,
option $(A)$ is correct answer.