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3 years ago

Anyone tell me what is the physics

Physics

1 answer:

Katyanochek1 [597]3 years ago

6 0

Answer:

physic is branch of science in which math is its brother and will deal with the equation,law and evidence of natural phenomenon.

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A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and

kari74 [83]

Answer:

$a. 4.2057\times 10^-^1^2 F \ or 4.2057\ pF\\b. 2.7344V$

Explanation:

Given the permittivity constant to be $8.854\times 10^-^1^2 F/m$ ,The capacitance of a $cylindrical \ capacitor$ of length, $L$ is given by the equation:

$C=\frac{2\pi \epsilon _o L}{ln(b/a)}$ where $b$ is the radius of the outer cylinder and $a$ the radius of the inner cylinder.

The values are given as: $a=0.550mm(5.5\times 10^-^4m), \ b=4.00mm(4.0\times10^-^3m), \ L=15.0cm(0.150m)$

Substitute in our capacitance equation:

$C=\frac{2\times\pi \times 8.854\times 10^-^1^2 \times 0.15}{In(4.00/0.550)}\\=4.2057\times 10^-^1^2 F$

Hence the capacitance is $4.2057\times 10^-^1^2 F$

b. The charge on the capacitance is related to the potential difference across it. The potential difference is expressed using the equation:

$Q=CV$ , $Q=11.5pC$

From a above, we already have our capacitance value, $C=4.2057\times 10^-^1^2 F$

We substitute $C$ in the pd equation:

$v=>(11.5)/(4.2057)\\=2.7344V$

Hence, the applied potential difference is 2.7344V

4 0

3 years ago

A. Two point charges totaling 7.50 µC exert a repulsive force of 0.100 N on one another when separated by 0.911 m. What is the c

andriy [413]

coulomb's law

f repulsive = q1q2/4 pi epsilon nought r squared

0.1= q1q2/4 pi epsilon nought 0.911 squared

q1+q2= 7.50 µC

0.1= q1(7.5µC-q1)/4 pi epsilon nought 0.911 squared

solve for q1

7 0

3 years ago

When an object with a negative charge is moved from point A to point B through an external electrical field, it gains electrical

Sergio039 [100]

Explanation:

As per the problem,

$\Delta U = (V_{B} - V_{A})(-q) > 0$

When q > 0 then -q is a negative charge . Since, change in potential energy ( $\Delta U$ ) increases.

or, $(V_{A} - V_{B})q$ > 0

or, $V_{A} > V_{B}$

Therefore, both positive and negative charge will move from $V_{A}$ to $V_{B}$ and as $V_{B} < V_{A}$ so both of them move through a negative potential difference.