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mina [271]
3 years ago
13

A charge of 50 µC is pushed by a force of 25 µN a distance of 15 m in an electric field. What is the electric potential differen

ce? Recall that Fe = qE.
Physics
2 answers:
Natalka [10]3 years ago
5 0

Answer:

ΔV = 7.5 V

Explanation:

Irina-Kira [14]3 years ago
4 0
Electric potential differnce is defined as:
\Delta V=\int_{a}^{b}E\cdot dl
If we asume the electric field is constant we get:
\Delta V=\int_{a}^{b}E\cdot dl=E\Delta L
We need to calculate the electic field strenght:
F=qE\\


E= F/q\\
E=\frac{25\cdot 10^{-6}}{50 \cdot 10^{-6}}=0.5 \frac{V}{m}
We can now calculate the potential difference:
\Delta V=E\Delta L=0.5 \frac{V}{m}\cdot15m=7.5V
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Explanation:

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3 years ago
A sound wave from a police siren has an intensity of 100.0 W/m² at a certain point; a second sound wave from a nearby ambulance
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<h3>What do you mean by sound?</h3>

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2 years ago
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Gwar [14]
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