Q10. Three points A, B and C lie in uniform electric field (E) of 5 x 10^3 N/C such that AC=5cm, BC=3cm. A, B and c form a right triangle such that AB is parallel to direction of E, BC is perpendicular to the direction of E, ∠B =90°. Find the potential difference between A and C.
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Potential difference is the dot product of electric field intensity and speration vector. e.g., 
Where , ∆V is potential difference,
E is electric field intensity,
and ∆r is separation vector.
A rough daigram is shown in figure.
Here it is clear that ,
∆V = |E| |∆r| cosФ
Ф is the angle between Electric field E and separation vector ,∆r
See attachment, cosФ = AB/AC = 4/5
|E| = 5 × 10³ N/C, ∆r = AC = 5cm = 0.05 m
∴ ∆V = 5 × 10³ × 0.05 × 4/5
= 0.2 × 10³ = 200 V
Hence answer is 200V
Where , ∆V is potential difference,
E is electric field intensity,
and ∆r is separation vector.
A rough daigram is shown in figure.
Here it is clear that ,
∆V = |E| |∆r| cosФ
Ф is the angle between Electric field E and separation vector ,∆r
See attachment, cosФ = AB/AC = 4/5
|E| = 5 × 10³ N/C, ∆r = AC = 5cm = 0.05 m
∴ ∆V = 5 × 10³ × 0.05 × 4/5
= 0.2 × 10³ = 200 V
Hence answer is 200V
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