Respuesta :
Answer:
(A). The work done is [tex]1.50\times10^{-6}\ J[/tex].
(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 [tex]K.E=1.50\times10^{-6}\ J[/tex]
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
[tex]W = \Delta K.E[/tex]
[tex]W=K.E_{f}-K.E_{i}[/tex]
Put the value into the formula
[tex]W= 1.50\times10^{-6}-0[/tex]
[tex]W=1.50\times10^{-6}\ J[/tex]
The work done is [tex]1.50\times10^{-6}\ J[/tex].
(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
[tex]\Delta P.E=\Delta K.E[/tex]
So, [tex]U = 1.50\times10^{-6}[/tex]
Using formula of potential
[tex]V=\dfrac{U}{q}[/tex]
Put the value into the formula
[tex]V=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}}[/tex]
[tex]V=357.14\ V[/tex]
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
[tex]W=F\times r[/tex]....(I)
Using formula of force
[tex]F=qE[/tex]
Put the value in the equation (I)
[tex]W=qE\times r[/tex]
[tex]E=\dfrac{W}{q\times r}[/tex]
Put the value into the formula
[tex]E=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}\times6.00\times10^{-2}}[/tex]
[tex]E=5952.38\ N/C[/tex]
The magnitude of E is 5952.38 N/C.
Hence, This is the required solution.
Part A: The work done by the charged particle is [tex]1.50\times 10^{-6}[/tex] Joules.
Part B: The potential difference between starting and end points is 357.14 V.
Part C: The magnitude of E on the charged particle is 5952.38 N/C.
How do you calculate the work and potential difference?
Given that the charge of the particle is +4.20nC and the kinetic energy is +1.50×10^−6J. The distance traveled by the charge is 6 cm. The charged particle is released from the rest, hence its initial velocity is zero.
Part A
The work done by the charged particle is calculated by its kinetic energy.
[tex]W = \Delta KE[/tex]
[tex]W = KE_f - KE_i[/tex]
Where KE_f is the final kinetic energy and KE_i is the initial kinetic energy. But the charged particle is released from rest, hence its initial kinetic energy will be zero.
[tex]W = KE_f[/tex]
[tex]W= 1.50 \times 10^{-6} \;\rm J[/tex]
Hence we can conclude that the work done by the charged particle is [tex]1.50\times 10^{-6}[/tex] Joules.
Part B
The potential difference between starting and ending points is calculated by the work done by the charged particle.
[tex]\Delta V = \dfrac {W}{q}[/tex]
Here [tex]\Delta V[/tex] is the potential difference between starting and end points, and q is the charge on the particle.
[tex]\Delta V = \dfrac {1.5 \times 10^{-6}}{4.20 \times 10^{-9}}[/tex]
[tex]\Delta V = 357.14 \;\rm V[/tex]
Hence we can conclude that the potential difference between starting and end points is 357.14 V.
Part C
The magnitude of E is calculated as given below.
[tex]W = F\times r[/tex]
Where W is work done, r is the distance traveled and F is the force on the charged particle.
[tex]W = qE \times r[/tex]
[tex]E = \dfrac { W}{q\times r}[/tex]
Substituting the values.
[tex]E = \dfrac {1.5\times 10^{-6}}{4.20\times 10^{-9}\times 6\times 10^{-2}}[/tex]
[tex]E = 5952.38 \;\rm N/C[/tex]
Hence we can conclude that the magnitude of E on the charged particle is 5952.38 N/C.
To know more about the work and potential difference, follow the link given below.
https://brainly.com/question/5630757.
