This is defined as the electric force experienced per unit charge. It is a vector quantity and is expressed in N/C.
Electric Potential VE
The electric potential VE at a point in an electric field is defined as the work done in taking a unit charge from infinity to that point. It is a scalar quantity and is expressed in volt.
Calculate the electrostatic force that exists between two charges of values 5.3μC and -3.2μC placed at 6.0 × 10-4m. (k = 9 × 109Nm2/C2)
Determine the electric potential due to charge 5.3μC placed at that point.
Three charges +10C, -20C and 16C are distributed as shown in the figure below. Find the resultant force acting on the 10C charge.
Let the Force of attraction between +10C and – 20C be
Let the Force of attraction between +10C and – 16C be
Net force = (F21+F22)−−−−−−−−√=(2×1011)+(3.6×1011)2−−−−−−−−−−−−−−−−−−−−√=4.2×1011NTanθ=F1F2=2×10113.6×1011=0.555θ=tan−10.555=32.28o
The net force is 4.12 × 1011N in a direction 32.280 to the line joining the +10C and – 16C charges.
Two similar but opposite point charges –q and +q each of magnitude 5.0 x 10 –8 C are separated by a distance of 8.0cm in a vacuum as shown below.
Calculate the magnitude and direction of the resultant electric field intensity Eat the point p.
At p the direction of the field vector is towards the left since the positive charge at p is attracted towards the –q charge.
The field vector due to +q charge is also directed towards the left since the positive charge at p is repelled by +q
Resultant electric field is E=E−+E+
The field is directed towards the left or towards –q charge.
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