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Jet001 [13]
2 years ago
10

What is the electric potential energy of a 4. 0 uc charge placed at that point where the electric potential at that point in spa

ce is 8 v?
Physics
2 answers:
dexar [7]2 years ago
6 0

The electric potential at that point in space is 8 v is 32uj

what is electric potential energy?

Potential energy is the energy within an object relative to its position and proximity to other objects within a field.

The electric potential of a point charge is V=kQ/rV=kQ/r , or V = k Q / rV=kQ/r . Electric potential is a scalar while the electric field is a vector. While the total electric field can be computed by adding individual fields as vectors, the voltage resulting from a collection of point charges can be calculated by adding voltages as integers.

It is decided to set the potential at infinity to zero.

but V for a point charge reduces with distance while decreasing with distance squared. Remember that the electric field E is a vector, but the electric potential V is a scalar with no direction.

The electric energy is (8×4)=32 uj

learn more about electric potential from here: https//brainly.com/question/28167853

#SPJ4

nexus9112 [7]2 years ago
3 0

The electric energy is 32 uj

The electric potential of a point charge is V=kQ/r, or V = k Q / r. Electric potential is a scalar while the electric field is a vector. While the total electric field can be computed by adding individual fields as vectors, the voltage resulting from a collection of point charges can be calculated by adding voltages as integers.

It is decided to set the potential at infinity to zero. E=Fq=kQr^2, but V for a point charge reduces with distance while decreasing with distance squared. Remember that the electric field E is a vector, but the electric potential V is a scalar with no direction.

The electric energy is (8×4)=32 uj

Learn more about electric energy  herebrainly.com/question/24403138

#4219.

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Explanation:

a) d\Phi = B.dA\\\\B_r = \frac{\mu_0I}{2\pi r}\\\\dA = b.dr\\\\\Phi = \int\limits^b_a {B.dA} = \int\limits^{a+d}_d {\frac{\mu_0I}{2\pi r}.b.dr} =\frac{\mu_0I}{2\pi }.b.ln(\frac{a+d}{d} )\\

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\Phi = \frac{4\pi * 10^{-7}*16.15*0.027}{2\pi} ln(1.46/0.46) = 1.007 * 10^{-7} V.m

Direction is into the page.

b) emf = -d\Phi/dt = -\frac{\mu_0}{2\pi }.b.ln(\frac{a+d}{d}).(2.98t)= \\=-\frac{4\pi * 10^{-7}*0.027}{2\pi} ln(1.46/0.46)*2.98*2.90= -5.39 * 10^{-8} V

Direction is counterclockwise.

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