I don't know how good you are at sketching ... I'm terrible.
But you can put the point across in a dramatic way if you
can sketch a bowling ball and a basketball ... you'll need
to clearly identify them with the markings you sketch on
each ball.
They're the same shape and nearly the same size, but
there's a huge difference in their densities.
Answer:
![\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%5Ctimes%28pie%29%5Ctimes%5Ctext%7BE%7D%7D%20%5Ctimes%5Cfrac%7Bq%7D%7BI%5E%7B2%7D%20%7D%2B%5Cfrac%7B1%7D%7B4%5Ctimes%28pie%29%5Ctimes%5Ctext%7BE%7D%7D%20%5Ctimes%5Cfrac%7B-q%7D%7BI%5E%7B2%7D%20%7D)
Explanation:
As given point p is equidistant from both the charges
It must be in the middle of both the charges
Assuming all 3 points lie on the same line
Electric Field due a charge q at a point ,distance r away
![=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B4%5Ctimes%28pie%29%5Ctimes%5Ctext%7BE%7D%7D%20%5Ctimes%5Cfrac%7Bq%7D%7Br%5E%7B2%7D%20%7D)
Where
- q is the charge
- r is the distance
-
is the permittivity of medium
Let electric field due to charge q be F1 and -q be F2
I is the distance of P from q and also from charge -q
⇒
F1![=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B4%5Ctimes%28pie%29%5Ctimes%5Ctext%7BE%7D%7D%20%5Ctimes%5Cfrac%7Bq%7D%7BI%5E%7B2%7D%20%7D)
F2![=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B4%5Ctimes%28pie%29%5Ctimes%5Ctext%7BE%7D%7D%20%5Ctimes%5Cfrac%7B-q%7D%7BI%5E%7B2%7D%20%7D)
⇒
F1+F2=![\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%5Ctimes%28pie%29%5Ctimes%5Ctext%7BE%7D%7D%20%5Ctimes%5Cfrac%7Bq%7D%7BI%5E%7B2%7D%20%7D%2B%5Cfrac%7B1%7D%7B4%5Ctimes%28pie%29%5Ctimes%5Ctext%7BE%7D%7D%20%5Ctimes%5Cfrac%7B-q%7D%7BI%5E%7B2%7D%20%7D)
Gravitational potential energy can be calculated using the formula <span>PE = m × g × h, where g is the gravitational acceleration and is constant hence the energy is dependent directly to mass and the height of the object. Hence more PE is registered when the object is heavier and/or at greater initial height. </span>
The answer is X. I hope this helped
Answer:
EP = 49.05Joules (J)
Explanation:
The equation for Potential energy (EP) is
EP = m g h
We are given the values below (do convert them into SI units)
m = 0.0025kg
h = 2000m
g = 9.81m/
Substitute the values into the equation and solve for EP
EP = 0.0025 * 2000 * 9.81
EP = 49.05Joules (J)