1. Why is an electric field considered to be a vector quantity?
2. Draw the electric field around an electric dipole of +1 µC and -1 µC
3. How many electrons have been removed from a positively charged object if it has a charge of 1.5 x 1011 C?
4. A particle with a charge - 1.22 x 10-7 C is at a point where the electric field is 4.00 x 103 N/C directed to the right. What is the magnitude and direction of the electric force acting on the particle?
5. A positively-charged oil drop weight 4.01 x 10-14 N. The drop is suspended in an electric field intensity of 7.6 x 103 N/C.
a) What is the charge on the oil drop?
b) How many electrons is the oil drop missing?
6. A positive test charge of + 6.5 x 10 -6 C experiences a force of +4.5 x 10 -5 N. What is the magnitude of the Electric field intensity ?
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What is it you do not understand about the question?
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1. An electric field is considered to be a vector quantity because it has both magnitude and direction. It describes the force experienced by a positive test charge at a given point in space. The magnitude of the electric field represents the intensity or strength of the field, while the direction indicates the direction in which a positive test charge would experience a force.
To determine the electric field at a given point, you can use Coulomb's law or the principle of superposition. Coulomb's law states that the electric field at a point due to a single charge is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance from the charge. The direction of the electric field is from positive charges toward negative charges.
2. To draw the electric field around an electric dipole, we need to consider the individual electric fields created by each charge. A dipole consists of two equal and opposite charges separated by a distance. In this case, we have a +1 µC charge and a -1 µC charge.
We can start by drawing two arrows to represent the electric fields created by each charge. The arrows should point away from the positive charge and toward the negative charge. The length of the arrows represents the magnitude of the electric field.
Once you have drawn the individual electric fields, draw an additional arrow at the midpoint between the charges to represent the net electric field. The net electric field at this point is the vector sum of the individual electric fields created by the charges.
3. To determine the number of electrons removed from a positively charged object, you need to know the charge of an electron and the given charge of the object.
The elementary charge, represented by the symbol "e," is equal to 1.6 x 10^-19 C. One electron has a charge of -1e. So, to find the number of electrons removed, divide the given charge of the object by the charge of one electron.
In this case, the charge of the object is 1.5 x 10^11 C. Therefore, divide this value by the charge of one electron to find the number of electrons removed.
4. The electric force acting on a charged particle can be determined using the equation F = qE, where F is the electric force, q is the charge of the particle, and E is the electric field.
In this case, the charge of the particle is -1.22 x 10^-7 C, and the electric field is 4.00 x 10^3 N/C directed to the right. The charge is negative, so the force will be in the opposite direction to the electric field.
To find the magnitude of the electric force, multiply the charge of the particle by the magnitude of the electric field. Then, determine the direction of the force based on the charges of the particle (negative) and the direction of the electric field.
5a. To determine the charge on the oil drop, we can use the equation F = qE, where F is the weight of the drop, q is the charge on the drop, and E is the electric field intensity.
In this case, the weight of the oil drop is given as 4.01 x 10^-14 N, and the electric field intensity is 7.6 x 10^3 N/C. Set up the equation F = qE and solve for q.
5b. To determine the number of electrons missing from the oil drop (assuming it carries a negative charge), we need to know the charge on a single electron.
The elementary charge, represented by the symbol "e," is equal to 1.6 x 10^-19 C. One electron has a charge of -1e. Divide the charge on the oil drop by the charge of one electron to find the number of missing electrons.
6. The magnitude of the electric field intensity can be determined using the equation E = F/q, where E is the electric field intensity, F is the electric force, and q is the charge of the test charge.
In this case, the electric force acting on the test charge is given as +4.5 x 10^-5 N, and the charge of the test charge is +6.5 x 10^-6 C. Divide the magnitude of the electric force by the charge of the test charge to find the magnitude of the electric field intensity.