Answer:
13.4 x 10 raise to power -19 C
Explanation:
. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m
. The uniform electric field is E = 214 N/M
, The decrease in electrical potential energy is
d(P.E) = 51.63 x 10 raise to power -19 J
Let the magnitude of the charge of the moving particle be q
which is given by the equation
d(P.E) =qEd
51.63 x 10 power -19 = q(214)(0.018)
51.63 x 10 power -19 =3.852q
by making q the formular,
q = 13.4 x 10 power -19 C
<h2>
Answer:</h2>
(a) 3.96 x 10⁵C
(b) 4.752 x 10⁶ J
<h2>
Explanation:</h2>
(a) The given charge (Q) is 110 A·h (ampere hour)
Converting this to A·s (ampere second) gives the number of coulombs the charge represents. This is done as follows;
=> Q = 110A·h
=> Q = 110 x 1A x 1h [1 hour = 3600 seconds]
=> Q = 110 x A x 3600s
=> Q = 396000A·s
=> Q = 3.96 x 10⁵A·s = 3.96 x 10⁵C
Therefore, the number of coulombs of charge is 3.96 x 10⁵C
(b) The energy (E) involved in the process is given by;
E = Q x V -----------------(i)
Where;
Q = magnitude of the charge = 3.96 x 10⁵C
V = electric potential = 12V
Substitute these values into equation (i) as follows;
E = 3.96 x 10⁵ x 12
E = 47.52 x 10⁵ J
E = 4.752 x 10⁶ J
Therefore, the amount of energy involved is 4.752 x 10⁶ J
The answer would be B, George Darwin :)
