Physics, 18.07.2019 05:00 babsst7480

Acharge q of 1.3 × 10-16 coulombs moves from point a to a lower potential at point b in an electric field of 3.2 × 102 newtons/coulomb. if the distance traveled is parallel to the field is 1.1 × 10-2 meters, what is the difference in the potential energy? 1.22 × 10-15 joules -2.4 × 10-15 joules -3.2 × 10-15 joules -4.6 × 10-16 joules 5.6 × 10-15 joules

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Answer from: Jamilia561

Difference in potential energy is, $\Delta V=-4.6\times 10^{-16}$

Explanation:

It is given that,

Charge, $q=1.3\times 10^{-16}\ C$

It moves from potential A to a lower potential B

Electric field, $E=3.2\times 10^2\ N/C$

Distance travelled, $r=1.1\times 10^{-2}\ m$

We have to find the difference in potential energy. We know that the work done is stored in the form of potential energy.

Also, $\Delta V=\dfrac{\Delta U}{q}$

$\Delta U$ is potential energy

And $E=-\dfrac{dV}{dr}$

So, W=-qEr

$W=-1.3\times 10^{-16}\times 3.2\times 10^2\times 1.1\times 10^{-2}$

$W=-4.576\times 10^{-16}\ J$

$W=-4.6\times 10^{-16}\ J$

Hence, the difference in potential energy is "-4.6 × 10⁻¹⁶ J"

Answer from: Quest

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Answer from: Quest

Idon't know it i just need my points lol

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Acharge q of 1.3 × 10-16 coulombs moves from point a to a lower potential at point b in an electric...

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