(a) Find the acceleration of B.<br>(b) Find the tensions, T1 and T2, in the strings.

There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

taurus [48]

Answer:

The total electric potential at mid way due to 'q' is $\frac{q}{4\pi\epsilon_{o}d}$

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say) = d

The electric potential at a distance d due to 'Q' is:

$V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}$

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

$V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Similar is the case with plate B:

$V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

$V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}$

$V_{total} = \frac{q}{4\pi\epsilon_{o}d}$

Now,

The Electric field due to charge Q at a distance is given by:

$\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}$

Now, if the charge q is mid way between the field, then distance is $\frac{d}{2}$ .

Electric Field at plate A, $\vec{E_{A}}$ at midway due to charge q:

$\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Similarly, for plate B:

$\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

3 0

2 years ago

A model rocket is launched directly upward at a speed of 16 meters per second from a height of 2 meters. The function f(t)=−4.9t

Crank

Answer:

$Hmax=15.06$ meters

Explanation:The question ask for the maximum value of the function f(t) which can be find by find the maxima of the function

The maxima of the function occurs when the slope is zero. i.e.

$\frac{df}{dt} =0\\\frac{df}{dt} =\frac{d}{dt} (-4.9t^2+16t+2)\\\frac{df}{dt} =-4.9*2t+16\\-9.8t+16=0\\t=16/9.8\\t=1.63 secs$

Hence the maxima occurs at t=1.63 seconds

The maximum value of f is

$f(1.63)=-4.9(1.63^2)+16(1.63)+2\\f(1.63)=15.06\\$

hence maximum height is found to be

$Hmax=15.06$ meters

8 0

3 years ago

The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

kobusy [5.1K]

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

3 0

3 years ago

A 10 kg turkey, He kicks the 0.5 kg ball with a force of 50N for 0.2 seconds and the ball flies straight away horizontally from

Harman [31]

Answer:

a. 20m/s

b.50N

c. Turkey has a larger mass than the ball. Neglible final acceleration and therefore remains stationery.

Explanation:

a. Given the force as 50N, times as 0.2seconds and the weight of the ball as 0.5 kg, it's final velocity can be calculated as:

$F\bigtriangleup t=m\bigtriangleup v\\\\50N\times 0.2s=0.5kg\times \bigtriangleup v\\\\\bigtriangleup v=2(50N\times0.2)\\\\=20m/s$

Hence, the velocity of the ball after the kick is 20m/s

b.The force felt by the turkey:

#Applying Newton's 3rd Law of motion, opposite and equal reaction:

-The turkey felt a force of 50N but in the opposite direction to the same force felt by the ball.

c. Using the law of momentum conservation:

-Due to ther external forces exerted on the turkey, it remains stationery.

-The turkey has a larger mass than the ball. It will therefore have a negligible acceleration if any and thus remains stationery.

-Momentum is not conserved due to these external forces.

5 0

3 years ago

Please Help Meee

g100num [7]

Answer:

answer is 24 ohm okkkkkm

4 0

2 years ago

(a) Find the acceleration of B.(b) Find the tensions, T1 and T2, in the strings.