Physics, 09.11.2020 16:40, MyaMya12

Determine the rate at which the electric field changes between the round plates of a capacitor, 8.0 cm in diameter, if the plates are spaced 1.4 mm apart and the voltage across them is changing at a rate of 110 V/s .

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answered: amelvin41

Solution:

The relation between the potential difference and the electric field between the plates of the parallel plate capacitor is given by :

$E=\frac{V}{D}$

Differentiating on both the sides with respect to time, we get

$\frac{dE}{dt}=\frac{1}{D}\frac{dV}{dt}$

Therefore, the rate of the electric field changes between the plates of the parallel plate capacitor is given by :

$\frac{dE}{dt}=\frac{1}{D}\frac{dV}{dt}$

$=\frac{1}{1.4 \times 10^{-3}} \times 110$

$=7.85 \times 10^4$ V/m-s

answered: Guest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.

$_a^z\textrm{x}\rightarrow _{z-1}^a\textrm{y}+_{+1}^0e$

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

answered: Guest

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

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Determine the rate at which the electric field changes between the round plates of a capacitor, 8.0...