(i) The total capacitance for the circuit is 5 μF.
(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.
(iii) The charge stored in 3μF capacitor is 6 x 10⁻⁶ C.
<h3>Total capacitance of the circuit</h3>
The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.
<h3>C1 and C2 are in series </h3>

<h3>C1 and C2 are parallel to C3</h3>

<h3>C(123) is series to C5 and C6</h3>

<h3>C7 and C8 are in series</h3>

<h3>Total capaciatnce of the circuit</h3>
Ct + C(78) = 2 μF + 3 μF = 5 μF
<h3 /><h3>Total charge stored in the circuit</h3>
The total charge stored in the capacitor is calculated as follows;
Q = CV
Q = (5 x 10⁻⁶) x (20)
Q = 1 x 10⁻⁴ C
<h3>Charge stored in 3μF capacitor</h3>
Q = (3 x 10⁻⁶) x (20)
Q = 6 x 10⁻⁶ C
Learn more about capacitance of capacitor here: brainly.com/question/13578522
Answer:
no ... hahahha! but I know every boys wait for the day when their heart beat is faster than normal ever in life
The answer is: "
44
km " ;
or; write as: "
44.333 km " .
___________________________________________________________Explanation:___________________________________________________________(70 km + 63 km) ÷ (2 + 1 ) = 133 km ÷ 3 = "
44
km " ;
or; write as: "
44.333 km " .
___________________________________________________________
Answer:

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

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
F2
⇒
F1+F2=