An infinitely long thin straight wire with uniform linear charge density is shown in figure. Draw a Gaussian surface in order to calculate the electric direction field at P and mark direction of electric field at this point.

In the given question we have to use the concept of gauss law and the concept of gaussian surface and apply it to solve the given question. Hence, above will be the correct answer of the given question.

An infinitely long thin straight wire with uniform linear charge density is shown in figure. b Derive an expression to calculate electric field at this point P .

Consider a uniformly charged wire of infinite length having a constant linear charge density λ (Charge per unit length). Let P be a point at a distance r from the wire and E be the electric field at the point P. A cylinder of length l, radius r, closed at each end by plane caps normal to the axis is chosen as Gaussian surface. Consider a very small area ds on the Gaussian surface. By symmetry, the magnitude of the electric field will be the same at all points on the curved surface of the cylinder and directed radially outward. E and ds are along the same direction. The electric flux (ϕ) through curved surface = ∮ E d s cos θ ϕ = ∮ E d s [ ∵ θ = 0 ; cos θ = 1 ] = E ( 2 r π l ) [The surface area of the curved part is ] since E and ds are right angles 2πrl to each other, the electric flux through the plane caps =0. &there4; Total flux through the Gaussian surface, ϕ = E ( 2 π r l ) . The net charge enclosed by Gaussian surface is, q = λ l &there4; By Gauss's law, = E ( 2 π r l ) λ l ε 0 o r E = λ 2 π ε 0 r The direction of electric field E is radially outward, if line charge is positive and inward, if the line charge is negative.

A spherical shell of radius R is uniformly charged with charge + q . By Gauss's theorem, find the electric field intensity at a point p . Inside the spherical shell.

Field Due To A Uniformly Charged Thin Spherical Shell: Consider a uniformly charged hollow spherical conductor of radius R . Let q be the total charge on the surface. To find the electric field at P (at a distance r from the centre), we imagine a Gaussian spherical surface having radius r . Then, according to Gauss’s theorem we can write, ∫ E → . d s → = 1 ε 0 q The electric field is constant, at a distance T . So we can write, E ∫ ds = 1 ε 0 q E 4 πr 2 = 1 ε 0 q E = 1 4 πε 0 q r 2 Inside the spherical shell E = 0 When a spherical shell is charged, the charges get distributed uniformly over its outer surface and the charge inside the shell is zero. According to Gauss’s law, as the charge inside is zero, the electric flux at any point inside the shell will be zero. Obviously the electric field (electric flux per unit area) is also zero. Hence, it is the required answer.

A spherical shell of radius R is uniformly charged with charge + q . By Gauss's theorem, find the electric field intensity at a point p . Outside the spherical shell.

Filed due to a uniformly charged thin spherical shell: Consider a uniformly charged hollow spherical conductor of radius R . Let q be the total charge on the surface. To find the electric field at P (at a distance r from the centre), we imagine a Gaussian spherical surface having radius r . Then, according to Gauss’s theorem we can write, ∫ E → . d s → = 1 ε 0 q The electric field is constant, at a distance T . So we can write, E ∫ ds = 1 ε 0 q E 4 πr 2 = 1 ε 0 q E = 1 4 πε 0 q r 2 Hence, it is the required answer.

A closed surface encloses an electric dipole. What is the electric flux through the surface?

As we all know that electric flux depends on the net charge enclosed inside the body. The formula of flux is the product of 1 ε and charge. Here, net charge is zero. Therefore, total electric flux is zero.

The ratio of electric field on the equatorial point and at the axial point at equal distances from the centre of a short electric dipole is ______.

As, we all know that dipole is the product of charge and distance. Therefore, we can calculate the value of electric field using this expression in axial and equatorial point. Electric field at equatorial line is given as k p r 3 And at axial line is given as 2 k p r 3 Hence, the required value for blank is 1 : 2 .

An infinitely long thin straight wire with uniform linear charge density is shown in figure. b Derive an expression to calculate electric field at this point P .

Consider a uniformly charged wire of infinite length having a constant linear charge density λ (Charge per unit length). Let P be a point at a distance r from the wire and E be the electric field at the point P. A cylinder of length l, radius r, closed at each end by plane caps normal to the axis is chosen as Gaussian surface. Consider a very small area ds on the Gaussian surface. By symmetry, the magnitude of the electric field will be the same at all points on the curved surface of the cylinder and directed radially outward. E and ds are along the same direction. The electric flux (ϕ) through curved surface = ∮ E d s cos θ ϕ = ∮ E d s [ ∵ θ = 0 ; cos θ = 1 ] = E ( 2 r π l ) [The surface area of the curved part is ] since E and ds are right angles 2πrl to each other, the electric flux through the plane caps =0. &there4; Total flux through the Gaussian surface, ϕ = E ( 2 π r l ) . The net charge enclosed by Gaussian surface is, q = λ l &there4; By Gauss's law, = E ( 2 π r l ) λ l ε 0 o r E = λ 2 π ε 0 r The direction of electric field E is radially outward, if line charge is positive and inward, if the line charge is negative.

An infinitely long thin straight wire with uniform linear charge density is shown in figure. Draw a Gaussian surface in order to calculate the electric direction field at P and mark direction of electric field at this point.

In the given question we have to use the concept of gauss law and the concept of gaussian surface and apply it to solve the given question. Hence, above will be the correct answer of the given question.

The electric field lines of a positive charge is as shown in figure: Give the sign of work done by the field in moving a small positive charge from ' Q ' to ' P ' .

In moving a small positive charge from the point Q to P , the work has to be done by an external agency against the electric field.Therefore, work done by the field is negative. Hence, the correct answer is negative.

Two spheres encloses charges as shown in figure: What is the ratio of electric flux through S 1 and S 2 ?

Given that, Two hollow concentric spheres S 1 and S 2 enclosing charges Q and 2 Q . Now, We have to find the ratio of electric flux through them. So, electric flux passing through sphere S1 φ 1 = Q ε 0 . And, electric flux passing through sphere S2 φ 2 = Q + 2 Q ε 0 = 3 Q ε 0 . &there4; Ratio of flux through S 1 and S 2 is, ⇒ φ 1 φ 2 = Q ε 0 3 Q ε 0 = 1 3 .