Question

please reply the answer with steps​

Answer 1

Question :

Figure shows equipotential surface associated with a uniform electric field.

Fnd the magnitude and direction of the electric field.

$\rule{200}2$

Before,going to Solve the problem,let's know about some Formula's related to this problem and Equipotential surfaces.

${\purple{\boxed{\large{\bold{Equipotential\:surfaces}}}}}$

It is the locus of all points having same potential.

➝Properties of equipotential surfaces:

• Field lines are always perpendicular to equipotential surfaces.
• Work done to move a charged particle on equipotential surface is always zero.
• Two equipotential surfaces can never intersect.

➝Relation between electric field and electric potential:

$\rm\triangle\:V=-E\triangle\:r\cos\theta$

The negative sign implies that potential decreases in the direction of electric field.

Solution :

Part -1 :

Magnitudes of Electric Field : 100 V/m

Direction : -ve x - axis

Part -2:

Magnitudes of Electric Field : 200 V/m

Direction : At angle 120° ,+ve axis in anticlockwise direction.

$\rule{200}2$

For step by step explanation refer to the attachments

Answer 2

Answer:

Question :

Figure shows equipotential surface associated with a uniform electric field.

Fnd the magnitude and direction of the electric field.

\rule{200}2

Before,going to Solve the problem,let's know about some Formula's related to this problem and Equipotential surfaces.

{\purple{\boxed{\large{\bold{Equipotential\:surfaces}}}}}

Equipotentialsurfaces

It is the locus of all points having same potential.

➝Properties of equipotential surfaces:

Field lines are always perpendicular to equipotential surfaces.

Work done to move a charged particle on equipotential surface is always zero.

Two equipotential surfaces can never intersect.

➝Relation between electric field and electric potential:

\rm\triangle\:V=-E\triangle\:r\cos\theta△V=−E△rcosθ

The negative sign implies that potential decreases in the direction of electric field.

Solution :

Part -1 :

Magnitudes of Electric Field : 100 V/m

Direction : -ve x - axis

Part -2:

Magnitudes of Electric Field : 200 V/m

Direction : At angle 120° ,+ve axis in anticlockwise direction.

\rule{200}2

For step by step explanation refer to the attachments