Question

Two point charges q₁ = 2µC and q₂ = 1µC are placed at distances b = 1 cm and a = 2cm from the origin on the y and x axes as shown in figure the electric field vector at point P(a, b) will subtend an angle q with x-axis given by-

• A. tanq = 1
• B. tanq = 2
• C. tan q = 3
• D. tan q = 4

Answer 1

Answer: (B)

tanq = 2

Answer 2

V_B – V_A = – $\int _ { \mathrm { A } } ^ { \mathrm { B } } \overrightarrow { \mathrm { E } } \cdot \mathrm { dr }$ = $\int _ { \mathrm { B } } ^ { \mathrm { A } } \overrightarrow { \mathrm { E } } \cdot \mathrm { dr }$

Now V_B=K–$\frac { \mathrm { kq } } { \mathrm { a } }$=–

\=– =