Question

The potential of a point B (-20 m, 30 m) taking

the potential of a point A (30 m, -20 m) to be zero

in an electric field ⃗ = 10xî – 20ĵ NC-1

is​

Answer 1

Therefore the potential of a point B (-20, 30) is 3500 volts.

Given : The potential at a point A (30 , -20) to be zero in an electric field, E = (10x i - 20 j)

To find : The potential at a point B(-20, 30).

solution : using formula,

$v_{BA}=v_B-v_A=-\int\limits^b_a{E}ds$

⇒$v_B$ - 0 = -$\int\limits^{(-20,30)}_{(30,-20)}(10x\hat{i}-20\hat{j}).(dx\hat{i}+dy\hat{j})$

= -$10\int\limits^{-20}_{30}xdx+20\int\limits^{30}_{-20}dy$

= -10 [(20² - 30²)/2] + 20 (30 + 20)

= - 5 × -500 + 1000

⇒2500 + 1000

= 3500 volts

Therefore the potential of a point B (-20, 30) is 3500 volts.

Answer 2

Answer:

v

BA

=v

B

−v

A

=−

a

∫

b

Eds

⇒v_Bv

B

- 0 = -\int\limits^{(-20,30)}_{(30,-20)}(10x\hat{i}-20\hat{j}).(dx\hat{i}+dy\hat{j})

(30,−20)

∫

(−20,30)

(10x

i

^

−20

j

^

).(dx

i

^

+dy

j

^

)

= -10\int\limits^{-20}_{30}xdx+20\int\limits^{30}_{-20}dy10

30

∫

−20

xdx+20

−20

∫

30

dy

= -10 [(20² - 30²)/2] + 20 (30 + 20)

= - 5 × -500 + 1000

⇒2500 + 1000

= 3500 volts

Therefore the potential of a point B (-20, 30) is 3500 volts.