Help me!!!!!!!!!!!!!!!!!!!

Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.

Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0

3 years ago

A little girl riding a train rolls a ball toward the back of the train at 1.25 m/s NE. The train is traveling at a velocity of 1

Lelechka [254]

Answer:

Answer: 2.70m/s NE

Explanation:

Just did it.

4 0

4 years ago

If the kinetic and potential energy in a system are equal, then the potential energy increases. What happens as a result?

Fittoniya [83]

Answer:

The Kinetic Energy decreases. The Total Energy stays the same

Explanation:

The TE stays the same, so if PE increases then KE will decrease.

7 0

3 years ago

An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal but opposite charge on its pla

dusya [7]

Answer:

C). $U_f = \frac{U_0}{2}$

Explanation:

As we know that capacitance of a given capacitor is

$C = \frac{\epsilon_0 A}{d}$

now we know that energy stored in the capacitor plates

$U_0 = \frac{Q^2}{2C}$

here if all the dimensions of the capacitor plate is doubled

then in that case

$C' = \frac{\epsilon_0 (4A)}{2d}$

here area becomes 4 times on doubling the radius and the distance between the plates also doubles

So new capacitance is now

$C' = 2C$

so capacitance is doubled

now the final energy stored between the plates of capacitor is given as

$U_f = \frac{Q^2}{2C'}$

so the final energy is

$U_f = \frac{Q^2}{4C}$

$U_f = \frac{U_0}{2}$

4 0

3 years ago

An electron in the n = 7 level of the hydrogen atom relaxes to a lower energy level, emitting light of 397 nm. what is the value

Dimas [21]

Answer:

$n_f=2$

Explanation:

It is given that,

Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})$