The Sl unit for speed is

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

4 0

3 years ago

Seven seconds after a brilliant flash of lightning, thunder shakes the house. How far was the lightning strike from the house? S

Zepler [3.9K]

Answer:

About two kilometers away

$\rm distance=2.401\ km$

Explanation:

Given:

The time gap between the light and sound to travel to the house, $t=7\ s$

<em>Since the clouds are formed in the troposphere region of the atmosphere which extends from 8 kilometers to 12 kilometers above the earth-surface and the velocity of light is 300000 kilometers per second so it is visible almost instantly, hence we neglect the time taken by the light to travel to the house from the clouds.</em>

<u>∴Distance between the lightning-strike and the house:</u>

$\rm distance=v\times t$

we have the speed of sound as: $v=343\ m.s^{-1}$

So,

$\rm distance=343\times 7$

$\rm distance=2401\ m$

$\rm distance=2.401\ km$

6 0

2 years ago

A parallel plate capacitor can store

Norma-Jean [14]

A parallel plate capacitor can store electric charge and
electrical energy, and if the plates are far enough apart,
you can store your lunch in there too.

3 0

3 years ago

Read 2 more answers

How do the amplitudes of a 120 decibel sound and a 100-decibel sound compare?

NemiM [27]

The 120 decibel sound has more amplitude than the 100 decibel sound.

In Physics, the relation between amplitude and intensity is that the intensity of the wave is directly proportional to the square of its amplitude.

4 0

2 years ago

Effieiency of simple machine is always less than 100% why

zysi [14]

Answer:

efficiency of a machine is less than 100% because some part is energy is utilized to overcome some opposing forces like friction which is wasted as heat ,sound energy etc

Explanation:

8 0

3 years ago