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)
Answer:
One bulb could go out and the strand will stay on.
Explanation:
In series circuit, there is only one path provided for the current to flow. So, all the lights are required to be in working condition, for the others to work. And if anyone light bulb goes out, the circuit will become incomplete and the rest of the strand will also go out. Because there is only one path for current flow which is now broken.
On the other hand, in parallel circuits, each light bulb has a separate connection with the source. Current path to each bulb is independent of the others. Therefore, if one bulb goes out, the rest of the strand will stay on.
So, the correct option is:
<u>One bulb could go out and the strand will stay on.</u>
I believe your answer would be
B. Slowing Down.