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elena-14-01-66 [18.8K]
3 years ago
15

A particle with a charge of +4.20nC is in a uniform electric field E⃗ directed to the left. It is released from rest and moves t

o the left; after it has moved 6.00 cm, its kinetic energy is found to be +1.50×10−6J.Part AWhat work was done by the electric force?Part BWhat is the potential of the starting point with respect to the endpoint?Part CWhat is the magnitude of E⃗ ?
Physics
1 answer:
Morgarella [4.7K]3 years ago
7 0

Answer:

(A). The work done is 1.50\times10^{-6}\ J.

(B). The potential of the starting point with respect to the endpoint is 357.14 V.

(C). The magnitude of E is 5952.38 N/C.

Explanation:

Given that,

Charge = 4.20 nC

Distance = 6.00 cm

Kinetic energy K.E=1.50\times10^{-6}\ J

The particle start from rest.

So, the initial kinetic energy i zero.

(A). We need to calculate the work by the electric force

Using formula of work done

W = \Delta K.E

W=K.E_{f}-K.E_{i}

Put the value into the formula

W= 1.50\times10^{-6}-0

W=1.50\times10^{-6}\ J

The work done is 1.50\times10^{-6}\ J.

(B). We need to calculate the potential of the starting point with respect to the endpoint

We know that.

Change in potential energy = change in kinetic energy

\Delta P.E=\Delta K.E

So, U = 1.50\times10^{-6}

Using formula of potential

V=\dfrac{U}{q}

Put the value into the formula

V=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}}

V=357.14\ V

The potential of the starting point with respect to the endpoint is 357.14 V.

(C). We need to calculate the magnitude of E

Using formula of work done

W=F\times r....(I)

Using formula of force

F=qE

Put the value in the equation (I)

W=qE\times r

E=\dfrac{W}{q\times r}

Put the value into the formula

E=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}\times6.00\times10^{-2}}

E=5952.38\ N/C

The magnitude of E is 5952.38 N/C.

Hence, This is the required solution.

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