How to find initial velocity in projectile motion problems when you are not given the vlocity?

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

3 years ago

Bird bones have air pockets in them to reduce their weight—this also gives them an average density significantly less thanthat o

snow_tiger [21]

Answer:

41.4 g

41.4 cm³

1.08695 g/cm³

Explanation:

$\rho$ = Density of water = 1 g/cm³

Mass of water displaced will be the difference of the

$m=45-3.6\\\Rightarrow m=41.4\ g$

Mass of water displaced is 41.4 g

Density is given by

$\rho=\dfrac{m}{v}\\\Rightarrow v=\dfrac{m}{\rho}\\\Rightarrow v=\dfrac{41.4}{1}\\\Rightarrow v=41.4\ cm^3$

So, volume of bone is 41.4 cm³

Average density of the bird is given by

$\rho=\dfrac{45}{41.4}\\\Rightarrow \rho=1.08695\ g/cm^3$

The average density is 1.08695 g/cm³

3 0

3 years ago

A 35 N force makes a 10 degree angle with the positive x-axis. What is the magnitude of the vertical component of the force?

Snowcat [4.5K]

Answer:

6.07 N

Explanation:

Given that,

Force, F = 35 N

It makes 10 degree angle with the positive x-axis.

We need to find the magnitude of the vertical component of the force. It can be given by :

$F_y=F\sin\theta\\\\=35\times \sin(10)\\\\=6.07\ N$

So, the magnitude of the vertical component of the force is 6.07 N.

5 0

3 years ago

Read 2 more answers

What are some forces that come in pairs?

Nikolay [14]

Answer:

ACTION REACTION FORCES

Explanation:

When there is an action frce there will be a reaction force

3 0

2 years ago

Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when sp

grandymaker [24]

Answer:

frequency of the sound = f = 1,030.3 Hz

phase difference = Φ = 229.09°

Explanation:

Step 1: Given data:

Xini = 0.540m

Xfin = 0.870m

v = 340m/s

Step 2: frequency of the sound (f)

f = v / λ

λ = Xfin - Xini = 0.870 - 0.540 = 0.33

f = 340 / 0.33

f = 1,030.3 Hz

Step 3: phase difference

phase difference = Φ

Φ = (2π/λ)*(Xini - λ) = (2π/0.33)* (0.540-0.33) = 19.04*0.21 = 3.9984

Φ = 3.9984 rad * (360°/2π rad)

Φ = 229.09°

Hope this helps!

5 0

3 years ago