Explanation:
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In this problem we have the electric field intensity E:
E = 6.5 ×
newtons/coulomb
We have the magnitude of the load:
q = 6.4 ×
coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 ×
meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 ×
)(6.5 ×
)(1.2 ×
)
PE = 5.0 x
joules
None of the options shown is correct.
Answer:
A. it's the only answer that makes sense. if I'm wrong sorry
Explanation:
It is given that,
Let Charge of
is taken from points A & B such that
.
We need to find the energy of charge. Electric potential is defined as the work done per unit of electric charge. So,

So, the energy of charge decreases by
. Hence, the correct option is (a).
Energy is calculated as power*time, so give the wattage of 1200 W (equivalent to 1200 Joules/second) and time of 30 seconds, multiplying these gives 36000 J or 36 kJ of electrical energy.
If electrical charge or current is needed: Power = voltage * current, so given the power of 1200 watts and voltage of 120 V, current is 1200 W / 120 V = 10 Amperes. Charge is calculated by multiplying 10 A*30 s = 300 C.