the electric field is given by `vec(E)=(vec(F))/(q_(0))` here the test charge `q_(0)` should be (a) Infinitesimally small and positive (b) Infintestimally small and negativeA. only aB. only bC. a (or) bD. neither a and b

the electric field is given by `vec(E)=(vec(F))/(q_(0))` here the test

1.	the electric field is given by `vec(E)=(vec(F))/(q_(0))` here the test charge `q_(0)` should be (a) Infinitesimally small and positive (b) Infintestimally small and negativeA. only aB. only bC. a (or) bD. neither a and b
Answer» Correct Answer - 4

Discussion

No Comment Found

Related InterviewSolutions

One-fourth of a sphere of radius R is removed as shown in figure. An electric field E exists parallel to the xy plane . Find the flux through the remaining curved part. A. `piR^(2)E`B. `sqrt(2)piR^(2)E`C. `pir^(2)E//sqrt(2)`D. `2piR^(2)E`
A cylinder of radius `R` and length `L` is placed in the uniform electric field `E` parallel to the cylinder axis.The total flux from the two flat surface of the cylinder is given byA. `2piR^(2)E`B. `(piR^(2))/(E)`C. `(piR^(2)-piR)/(E)`D. zero
A bob of a simple pendulum of mass `40gm` with a positive charge `4xx10^(-6)C` is oscillating with a time period `T_(1)`.An electric field of intensity `3.6xx10^(4)N//C` is applied vertically upwards.Now the time period is `T_(2)` the value of `(T_(2))/(T_(1))` is `(g=10//s^(2))`A. `0.16`B. `0.64`C. `1.25`D. `0.8`
The inward and outward electric flux for a closed surface unit of `N-m^(2)//C` are respectively `8xx10^(3)` and `4xx10^(3)`. Then the total charge inside the surface is [where `epsilon_(0)=` permittivity constant]A. `4xx10^(3)`B. `-4xx10^(3)`C. `-(piR^(2)-piR)/(E)`D. `-4xx10^(3)epsilon_(0)`
A hemispherical body of radius R is placed in a uniform electric field E. What is the flux linked with the curved surface if, the field is (a) parallel to the base, (b) perpendicular to the base.A. `2piR^(2)E`B. `piR^(2)E`C. `4piR^(2)E`D. `6piR^(2)E`
A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at `x=+1cm ` and C be the point on the y-axis at `y=+1`cm. then the potetial at the points A,B and C satisfya. `V_AltV_B`, b. `V_AgtV_B` c. `V_AltV_C` d. `V_AgtV_C`A. `V_(A)lt V_(B)`B. `V_(A)gt V_(B)`C. `V_(A)lt V_(C)`D. `V_(A)gt V_(C)`
A cube of arranged such that its length ,breadth ,height are along`X,Y,Z` directions one of its corners is situated at the origin Length of each side of the cube is `25cm` .The components of electric field are `E_(x)=400sqrt(2)N//C` E_(y)=0` and `E_(z)=0` respectively.Flux coming out of the cube at one end will beA. `25sqrt(2)Nm^(2)//C`B. `5sqrt(2)Nm^(2)//C`C. `250sqrt(2)Nm^(2)//c`D. `25Nm^(2)//C`
`A,B,C` are three points as a circle of radius `1cm`.These points form the corners an equilateral triangle .A charge `2C` is placed at the centre of circle .The work done in carrying a charge of `0.1muc` form `A` to `B` isA. zeroB. `18xx10^(11)J`C. `1.8xx10^(11)J`D. `54xx10^(11)J`
Point charge of `3xx10^(-9)C` are situated at each of three corners of a square whose side is `15cm`. The magnitude and direction of electric field at the vacant corner of the square isA. `2296V//m` along the diagonalB. `9622V//m` along the diagonalC. `22.0V//m` along the diagonalD. Zero
Three charges each`20muC` are placed at the corners of an equilateral triangle of side `0.4m` The potential energy of the system isA. `18xx10^(-6)J`B. `9J`C. `9xx10^(-6)J`D. `27J`

the electric field is given by `vec(E)=(vec(F))/(q_(0))` here the test charge `q_(0)` should be (a) Infinitesimally small and positive (b) Infintestimally small and negativeA. only aB. only bC. a (or) bD. neither a and b

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment