Answer:
![\vec{F}= -3.52\times 10^{-13}\hat{i}\ N](https://tex.z-dn.net/?f=%5Cvec%7BF%7D%3D%20-3.52%5Ctimes%2010%5E%7B-13%7D%5Chat%7Bi%7D%5C%20N)
Explanation:
given,
charge = -5.0 μC
Electric force, F = 11 i^ N
force would a proton experience = ?
we know
![\vec{F} = q \vec{E}](https://tex.z-dn.net/?f=%5Cvec%7BF%7D%20%3D%20q%20%5Cvec%7BE%7D)
![11\hat{i} = -5 \times 10^{-6} \vec{E}](https://tex.z-dn.net/?f=11%5Chat%7Bi%7D%20%3D%20-5%20%5Ctimes%2010%5E%7B-6%7D%20%5Cvec%7BE%7D)
![\vec{E} =-2.2 \times 10^{6}\hat{i}](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D-2.2%20%5Ctimes%2010%5E%7B6%7D%5Chat%7Bi%7D)
we know charge of proton is equal to 1.6 x 10⁻¹⁹ C
using formula
![\vec{F} = q \vec{E}](https://tex.z-dn.net/?f=%5Cvec%7BF%7D%20%3D%20q%20%5Cvec%7BE%7D)
![\vec{F}= 1.6 \times 10^{-19}\times -2.2 \times 10^{6}\hat{i}](https://tex.z-dn.net/?f=%5Cvec%7BF%7D%3D%201.6%20%5Ctimes%2010%5E%7B-19%7D%5Ctimes%20-2.2%20%5Ctimes%2010%5E%7B6%7D%5Chat%7Bi%7D)
![\vec{F}= -3.52\times 10^{-13}\hat{i}\ N](https://tex.z-dn.net/?f=%5Cvec%7BF%7D%3D%20-3.52%5Ctimes%2010%5E%7B-13%7D%5Chat%7Bi%7D%5C%20N)
Force experienced by the photon in the same field is equal to ![\vec{F}= -3.52\times 10^{-13}\hat{i}\ N](https://tex.z-dn.net/?f=%5Cvec%7BF%7D%3D%20-3.52%5Ctimes%2010%5E%7B-13%7D%5Chat%7Bi%7D%5C%20N)
Initial velocity (u) = 2 m/s
Acceleration (a) = 10 m/s^2
Time taken (t) = 4 s
Let the final velocity be v.
By using the equation,
v = u + at, we get
or, v = 2 + 10 × 4
or, v = 2 + 40
or, v = 42
The final velocity is 42 m/s.
Answer:
![9.96\cdot 10^{-10}J](https://tex.z-dn.net/?f=9.96%5Ccdot%2010%5E%7B-10%7DJ)
Explanation:
The capacitance of the parallel-plate capacitor is given by
![C=\epsilon_0 k \frac{A}{d}](https://tex.z-dn.net/?f=C%3D%5Cepsilon_0%20k%20%5Cfrac%7BA%7D%7Bd%7D)
where
ϵ0 = 8.85x10-12 C2/N.m2 is the vacuum permittivity
k = 3.00 is the dielectric constant
is the area of the plates
d = 9.00 mm = 0.009 m is the separation between the plates
Substituting,
![C=(8.85\cdot 10^{-12}F/m)(3.00 ) \frac{30.0\cdot 10^{-4} m^2}{0.009 m}=8.85\cdot 10^{-12} F](https://tex.z-dn.net/?f=C%3D%288.85%5Ccdot%2010%5E%7B-12%7DF%2Fm%29%283.00%20%29%20%5Cfrac%7B30.0%5Ccdot%2010%5E%7B-4%7D%20m%5E2%7D%7B0.009%20m%7D%3D8.85%5Ccdot%2010%5E%7B-12%7D%20F)
Now we can calculate the energy of the capacitor, given by:
![U=\frac{1}{2}CV^2](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B2%7DCV%5E2)
where
C is the capacitance
V = 15.0 V is the potential difference
Substituting,
![U=\frac{1}{2}(8.85\cdot 10^{-12}F)(15.0 V)^2=9.96\cdot 10^{-10}J](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B2%7D%288.85%5Ccdot%2010%5E%7B-12%7DF%29%2815.0%20V%29%5E2%3D9.96%5Ccdot%2010%5E%7B-10%7DJ)