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# Electric Field Intensity or Strength ε and Electric Potential VE

This is defined as the electric force experienced per unit charge. It is a vector quantity and is expressed in N/C.

Mathematically,

E=FCQ∴E=Q4πϵ0r2

## Electric Potential VE

The electric potential Vat 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.

Mathematically,

VE=Er

But E=Q4πϵ0r2VE=(Q4πϵ0r2)rVE=Q4πϵ0r

## Worked Examples

Example 1:

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)

Solution:

FC=kQ1Q2r2FC=9×109×5.3×10−6×3.2×10−6(6.0×10−4)2=4.24×105N

Example 2:

Determine the electric potential due to charge 5.3μC placed at that point.

Solution:

VE=Q4πϵ0r=5.3×10−6×9×1096×10−4=79,500KV

Example 3:

Three charges +10C, -20C and 16C are distributed as shown in the figure below. Find the resultant force acting on the 10C charge.

Solution:

Let the Force of attraction between +10C and – 20C be

F1=kQ1Q2r2F1=9×109×10×2032=2×1011N(k=14πϵ0r29×109)

Let the Force of attraction between +10C and – 16C be

F2=kQ1Q2r2F2=9×109×10×162×2=3.6×1011N(3.6×1011)2

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 32.28to the line joining the +10C and – 16C charges.

Example 4:

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.

k=14πϵ0r29×109Nm2C−2

Solution:

At p the direction of the field vector is towards the left since the positive charge at p is attracted towards the –q charge.

E−=−q4πϵ0r2=5.0×10−8×9×109(0.05)−2E−=1.8×105NC−1

The field vector due to +q charge is also directed towards the left since the positive charge at p is repelled by +q

E+=+q4πϵ0r2=5.0×10−8×9×109(0.03)−2E+=5×105NC−1

Resultant electric field is E=E−+E+

E=(1.8×105+5.0×105)NC−1

The field is directed towards the left or towards –q charge.