A uniform field of magnitude E $$=$$ $$2000$$ $$N/C$$ is directed θ$$=37$$∘ below the horizontal. Find the potential difference between P and R VP−VR (in$$ $$volts).

Question

A uniform field of magnitude E $$=$$ $$2000$$ $$N/C$$ is directed θ$$=37$$∘ below the horizontal. Find the potential difference between P and R VP−VR (in$$ $$volts).

Infinity Learn · Accepted Answer

The potential difference between two points P and R in electric field is, $V_{P} - V_{R} = - \int_{R}^{P} \vec{E} \cdot \vec{dl}$

As electric field is constant,

$V_{P} - V_{R} = - \int_{R}^{P} \vec{E} \cdot \vec{dl}$

$V_{P} - V_{R} = - \vec{E} \cdot \int_{R}^{P} \vec{dl} = - \vec{E} \cdot (\vec{r_{P}} - \vec{r_{R}})$

$\Rightarrow V_{P} - V_{R} = - \vec{E} \cdot Δ \vec{l} = - E \cdot Δl \cdot \cos (180^{\circ} - θ)$

Here, $Δl = PR = 5 cm = 5 \times 10^{- 2} m$

Hence, $V_{P} - V_{R} = - 2000 \times 5 \times 10^{- 2} (- \cos 37^{0})$

$= 2000 \times 5 \times 10^{- 2} \times \frac{4}{5} = 80 V$

Electrostatic potential

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A uniform field of magnitude E = 2000 N/C is directed $θ = 37^{\circ}$ below the horizontal. Find the potential difference between P and R $(V_{P} - V_{R})$ (in volts).

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Electrostatic potential

Remember concepts with our Masterclasses.

A uniform field of magnitude E = 2000 N/C is directed θ=37∘ below the horizontal. Find the potential difference between P and R VP−VR (in volts).

Ready to Test Your Skills?

free study material

A uniform field of magnitude E = 2000 N/C is directed $θ = 37^{\circ}$ below the horizontal. Find the potential difference between P and R $(V_{P} - V_{R})$ (in volts).