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2 years ago

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what is the capacitance of two square parallel plates 6.7 cm on a side that are separated by 1.8 mm of paraffin? the dielectric

constant of paraffin is 2.2.

1 answer:

almond37 [142]2 years ago

7 0

The capacitance of two square parallel plates will be 1.777 farads.

<h3>What is the capacitance of two parallel plates?</h3>

The body's capacity to hold electric charge is constrained by capacitance. There is a capacitance for each capacitor. Two metallic sheets of area A, spaced by d, make up the standard parallel-plate capacitor.

The parallel plate capacitor formula is given by:

C = ε Ad

Two square parallel plates 6.7 cm on a side are separated by 1.8 mm of paraffin. The dielectric constant of paraffin is 2.2. Then the capacitance of two square parallel plates will be given as,

C = ε Ad

C = 2.2 x 0.67 x 0.67 x 1.8

C = 1.777 farad

Two parallel square plates will have a capacitance of 1.777 farads.

More about the capacitance of two parallel plates link is given below.

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$v_m \approx -4.38\; \rm m \cdot s^{-1}$ (moving toward the incline.)

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If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this $m = 2.20\; \rm kg\!$ block right before the collision would also be approximately $77.695\; \rm J$ .

Calculate the velocity of that $m = 2.20\; \rm kg$ based on its kinetic energy:

$\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}$ .

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

$p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

$p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right before the collision: approximately $18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right after the collision: $(m\cdot v_m + m \cdot v_M)$ .

For momentum to conserve in this collision, $v_m$ and $v_M$ should ensure that $m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Kinetic energy of the two blocks right before the collision: approximately $77.695\; \rm J$ and $0\; \rm J$ . Sum of these two values: approximately $77.695\; \rm J\!$ .

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$\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right)$ .

Similarly, for kinetic energy to conserve in this collision, $v_m$ and $v_M$ should ensure that $\displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J$ .

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