Answer:
Option B. 5 nC
Explanation:
From the question given above, the following data were obtained:
Capicitance (C) = 100 pF
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:
1 pF = 1×10¯¹² F
Therefore,
100 pF = 100 pF × 1×10¯¹² F / 1 pF
100 pF = 1×10¯¹⁰ F
Next, we shall determine the quantity of charge. This can be obtained as follow:
Capicitance (C) = 1×10¯¹⁰ F
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Q = CV
Q = 1×10¯¹⁰ × 50
Q = 5×10¯⁹ C
Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:
1 C = 1×10⁹ nC
Therefore,
5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C
5×10¯⁹ C = 5 nC
Thus, the quantity of charge is 5 nC
The subordinate clause is "<span>who are loyal and industrious" and is used as an adjective to describe students.</span>
Its the answer of B speed and velocity
Answer:
Net Charge is -1.6 x 10 (to the negative 19th power) C. l
Explanation:
Answer:
C) 2.44 × 106 N/C
Explanation:
The electric flux through a circular loop of wire is given by
where
E is the electric field
A is the cross-sectional area
is the angle between the direction of the electric field and the normal to A
The flux is maximum when 
, so we are in this situation and therefore 
, so we can write
Here we have:
is the flux
d = 0.626 m is the diameter of the coil, so the radius is
r = 0.313 m
and so the area is
And so, we can find the magnitude of the electric field:
