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Bumek [7]
3 years ago
5

An electron and a positron are located 15 m away from each other and held fixed by some mechanism. The positron has the same mas

s and the same magnitude of charge as those of the electron, but its charge is positive. The electron and the positron are released at the same time by the mechanism. The electron and the positron begin to speed up towards each other. What velocities should they have when they are 2 m away from each other
Physics
1 answer:
ICE Princess25 [194]3 years ago
8 0

Answer:

Explanation:

Electrical potential energy will be converted into kinetic energy .

Electrical potential energy when distance was 15 m .

E₁ = 9 x 10⁹ x - q² /d where q is magnitude of charge on electron or positron

E₁ = 9 x 10⁹ x - ( 1.6 x 10⁻¹⁹ )² /15

= - 1.536 x 10⁻²⁹ J .

Electrical potential energy when distance was 2 m .

E₁ =9 x 10⁹ x - q² /d where q is magnitude of charge on electron or positron

E₁ = 9 x 10⁹ x - ( 1.6 x 10⁻¹⁹ )² /2

= -11.52 x 10⁻²⁹ J .

Decrease in energy = (11.52 - 1.536 ) x 10⁻²⁹

= 9.984 x 10⁻²⁹ J .

This energy will be converted into kinetic energy and they will be distributed equally in each .

Energy of each = 9.984 x 10⁻²⁹ /2

= 4.992 x 10⁻²⁹ J .

1/2 m v² = 4.992 x 10⁻²⁹ , m is mass of electron

.5 x 9.1 x 10⁻³¹ v² = 4.992 x 10⁻²⁹

v² = 109.71

v = 10.47 m/s .

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Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
3 years ago
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