AIEEE 2004 | Electrostatics Question 211 | Physics

A charge particle $$'q'$$ is shot towards another charged particle $$'Q'$$ which is fixed, with a speed $$'v'$$. It approaches $$'Q'$$ upto a closest distance $$r$$ and then returns. If $$q$$ were given a speed of $$'2v'$$ the closest distances of approaches would be

$$r/2$$

$$2r$$

$$r$$

$$r/4$$

AIEEE 2004

MCQ (Single Correct Answer)

-1

Four charges equal to -$$Q$$ are placed at the four corners of a square and a charge $$q$$ is at its center. If the system is in equilibrium the value of $$q$$ is

$$ - {Q \over 2}\left( {1 + 2\sqrt 2 } \right)$$

$${Q \over 4}\left( {1 + 2\sqrt 2 } \right)$$

$$ - {Q \over 4}\left( {1 + 2\sqrt 2 } \right)$$

$${Q \over 2}\left( {1 + 2\sqrt 2 } \right)$$

AIEEE 2004

MCQ (Single Correct Answer)

-1

A charged oil drop is suspended in a uniform field of $$3 \times {10^4}$$ $$v/m$$ so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge $$ = 9.9 \times {10^{ - 15}}\,\,kg$$ and $$g = 10\,m/{s^2}$$)

$$1.6 \times {10^{ - 18}}\,C$$

$$3.2 \times {10^{ - 18}}\,C$$

$$3.3 \times {10^{ - 18}}\,C$$

$$4.8 \times {10^{ - 18}}\,C$$

AIEEE 2003

MCQ (Single Correct Answer)

-1

If the electric flux entering and leaving an enclosed surface respectively is $${\phi _1}$$ and $${\phi _2},$$ the electric charge inside the surface will be

$$\left( {{\phi _2} - {\phi _1}} \right){\varepsilon _0}$$

$$\left( {{\phi _2} + {\phi _1}} \right)/{\varepsilon _0}$$

$$\left( {{\phi _2} - {\phi _1}} \right)/{\varepsilon _0}$$

$$\left( {{\phi _1} + {\phi _2}} \right){\varepsilon _0}$$

PREVIOUS NEXT

Questions Asked from Electrostatics (MCQ (Single Correct Answer))

Number in Brackets after Paper Indicates No. of Questions