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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.

EVALUATION (POST YOUR ANSWERS USING THE QUESTION BOX BELOW FOR EVALUATION AND DISCUSSION):

1. State two advantages of potentiometer over voltmeter.
2. Mention two precautions to be ensured in using the metre bridge.
3. State Coulomb’s law.
4. Define electric potential.

GENERAL EVALUATION (POST YOUR ANSWERS USING THE QUESTION BOX BELOW FOR EVALUATION AND DISCUSSION):

1. Mention an instrument used for comparing the emf of cells.
2. Write down the formulae for net resistance and emf in a circuit.
3. Define conductivity.
4. Write down the relationship between conductivity and resistivity.

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