Question

ABC is a right angled triangle situated in a uniform electric field $\overrightarrow E$ which is in the plane of the triangle. The points A and B are at the same potential of $15{\text{ }}V$ while the point C is at a potential of ${\text{20 }}V$ . AB = $3{\text{ cm}}$ and BC = ${\text{4 }}cm$. The magnitude of the electric field is (in S.I. Units)

A. $100$
B. $125$
C. $167$
D. $208$

Answer 1

Remember the concept of potential between two points and the concept of potential drop. Applying this concept along the sides of the given triangle, with the help of the given data and now we can easily solve this question. Remember the formula of potential difference ( $\Delta V = E\times d$ ) . This formula is enough to solve this problem.

Complete step by step answer:
Now let’s use the given values from the diagram, we can see that AB = $3{\text{ cm}}$ and BC = ${\text{4 }}cm$.Now see that it is given that A and B have the same potential, thus we can say AB is the equipotential Line. We do not need to calculate AC and use it as we require only parallel field lines and we have BC for that purpose.

Now, we know that the formula of potential difference is: $\Delta V = E \times d$ , where E is the electric field and d is the distance between the points of different potential. Thus, by applying the formula we get:
$E = \dfrac{{\Delta V}}{{d}}$
Substituting the given values,
$E = \dfrac{{(20 - 15)V}}{{4 \times {{10}^{ - 2}}m}}$
$\Rightarrow E = \dfrac{5\times 100}{4}$
Further simplifying we get,
$E = 5\times 25$
$\therefore E = 125$

Hence the value of the electric field has been determined and it is $125 V$.

Note: This question requires a basic concept of static electronics and can be easily solved using the formula directly. Now, you might wonder why we don’t consider the part AC, always remember that the field lines are parallel and thus we are to take the part BC. The most important thing is to understand the basic concepts of Electrostatics and electronics.