Untitled

9702_w22_qp_41-Q4

State what is indicated by the direction of an electric field line. (2)

• Direction of force on a positive charge

$$E=\frac{\Delta V}{\Delta d}$$

The helium nucleus has double the charge but four times the mass. Because there is no horizontal force, horizontal velocity of both particles is the same. Since vertical acceleration depends on charge-to-mass ratio, helium has a lower vertical acceleration. Therefore final speed of helium is lower.

explanation: horizontal velocity same as no horizontal resultant force, they spend the same time in the field vertical resultant force given by:

$$F=qE$$

Newton’s Second law:

$$F=ma$$ $$ma=qE$$ $$a=\frac{qE}{m}$$

For proton:

$$a=\frac{eE}{u}$$

For helium nucleus:

$$a=\frac{2eE}{4u}=\frac{1}{2}\left(\frac{eE}{u}\right)$$

Helium will have half the acceleration of the proton

9702_w25_qp_44-Q5

Explain why the electric potential near an isolated proton is positive

Electric potential is defined as zero at infinity. Because both a proton and a test charge are positively charged, there is a repulsive force between them. Therefore, work has to be done by an external force to bring the test charge from infinity to a point near the proton.

9702_w25_qp_42-Q6

Define electric field at a point Force per unit positive charge

9702_w25_qp_41-Q5

Define electric potential at a point

Work done per unit charge in brining a small positive test charge from infinity to the point.

9702_s25_qp_41-Q2

State Coulomb’s law The electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of their separation

9702_w24_qp_42-Q6

Suggest an explanation for the fact that the electric field inside a hollow sphere is zero.

Charges distribute evenly on the outer surface, therefore a test charge inside experiences equal and opposite forces in all directions, resulting in zero net electric field

9702_w24_qp_41-Q5

State the relationship between electric field and electric potential.

$$\text{Electric Field}=-\text{Potential Gradient}$$

Explain why it is not possible for the total electric potential and the resultant electric field to simultaneously be zero at point P

• For potential to be zero, the charges must have opposite signs
• For electric field to be zero, the charges must have the same sign.
• The signs of the charges cannot simultaneously be both the same and opposite

When fields in same direction, magnitudes add

9702_s24_qp_42-Q5

A proton is held at rest on the line joining the centres of the spheres at the position where x = 0.60 m. The proton is released. Describe and explain, without calculation, the subsequent motion of the proton.

• The proton remains at rest.
• There is no resultant electric force on the proton.

Explanation:

$$E=-\frac{dV}{dx}$$ $$E=0$$ $$F=qE$$ $$F=0$$

9702_w22_qp_42-Q5

Two identical negatively charge spheres are placed 0.46 m from each other. They are released simultaneously so that they are free to move. Describe and explain the subsequent motion of the spheres

Since they have the same sign of charge, they experience a repulsive force and move apart. The force is in the direction of motion so speed increases. As distance increases, electric force decreases so acceleration decreases. Momentum is conserved and masses are equal, so velocities are always equal and opposite.

9702_m23_qp_42-Q4

Explain why the electron does not follow a circular path. The force is not always perpendicular to the velocity as the force is always upwards.

9702_s23_qp_41-Q1

Assume that the Earth is a uniform conducting sphere of mass $5.98\times 10^{24}\text{ kg}$. The surface of the Earth carries a charge of $-4.80\times 10^5\text{ C}$ that is evenly distributed. State how the direction of the electric field at the surface of the Earth compares with the direction of the gravitational field. Same direction. Explanation: assuming the earth to be a uniform conducting sphere with charge evenly distributed, the sphere behave as both point charge and point mass. Gravitational field always attractive therefore points toward centre. Electric field for negative charge points toward centre. Both radial.

NOV 20 22 Q7

The electric field causes the electron to move from its initial position. Describe and explain the acceleration of the electron due to the field, as the electron moves through the field.

The acceleration of the electron increases. As the electron moves to the right, the electric field lines become closer, meaning that the electric field strength increases. The electric force on the electron increases. Since $a=\frac{F}{m}$, acceleration increases.

A stationary alpha particle is now placed in the same electric field at the same initial position that was occupied by the electron. Compare the initial electric force acting on the alpha particle with the initial force that acted on the electron.

The initial electric force on the $\alpha$-particle has twice the magnitude and acts in the opposite direction.

Explanation: electron has negative charge and alpha particle positive, so they will move in opposite directions electron has magnitude of charge $e$ alpha has $2e$ so alpha will experience twice the force

NOV 19 21 Q2

Explain why the resultant force on the bead is constant as the bead moves along path AB.

The electric force is constant as electric field strength of a uniform electric field is constant. The weight is constant as mass and gravitational field strength are constant. Resultant force is constant as vertical and horizontal components are constant.