Electric Forces and Fields

<p>Coulomb's constant (ke) (B)</p> Signup and view all the answers

<p>3x^2 + 8x - 8 = 0 (B)</p> Signup and view all the answers

<p>Equating the forces to zero (B)</p> Signup and view all the answers

<p>Factoring the quadratic equation (C)</p> Signup and view all the answers

<p>8 (C)</p> Signup and view all the answers

<p>Calculate the electric field from each charge separately. (A)</p> Signup and view all the answers

<p>The vector sum of the electric fields from each individual charge. (C)</p> Signup and view all the answers

<p>By finding the product of the total charge and the electric field. (C)</p> Signup and view all the answers

<p>Both the x- and y-components. (C)</p> Signup and view all the answers

<p>They create separate electric fields, which must be added vectorially. (C)</p> Signup and view all the answers

<p>Toward q2. (B)</p> Signup and view all the answers

<p>The angles related to the position of each charge must be used for accurate vector components. (D)</p> Signup and view all the answers

<p>It decreases the electric field strength. (D)</p> Signup and view all the answers

<p>The electric force F exerted by Q on q0. (C)</p> Signup and view all the answers

<p>Newton per coulomb (N/C) (D)</p> Signup and view all the answers

<p>To prevent any rearrangement of charge creating the electric field E. (B)</p> Signup and view all the answers

<p>It can be convenient but it is not rigorous. (A)</p> Signup and view all the answers

<p>It may alter the charge creating the electric field. (B)</p> Signup and view all the answers

<p>It is the source charge creating the electric field E. (B)</p> Signup and view all the answers

<p>F is calculated by multiplying E and q0. (A)</p> Signup and view all the answers

<p>E is the ratio of F to the magnitude of q0. (A)</p> Signup and view all the answers

<p>$5.42 imes 10^{-3} N$ (D)</p> Signup and view all the answers

<p>$5.84 imes 10^5 N/C$ (D)</p> Signup and view all the answers

<p>The electric field is stronger when lines are closer together. (D)</p> Signup and view all the answers

<p>The electric field vector E is tangent to the electric field lines at each point. (A)</p> Signup and view all the answers

<p>The force is in the same direction as the electric field. (A)</p> Signup and view all the answers

<p>Stronger electric fields in that region. (A)</p> Signup and view all the answers

<p>The problem should be addressed in small manageable steps. (C)</p> Signup and view all the answers

<p>He defined electric field lines to visualize electric field patterns. (C)</p> Signup and view all the answers

<p>3.93 $ 10^5 N/C (C)</p> Signup and view all the answers

<p>0 N/C (B)</p> Signup and view all the answers

<p>1.80 $ 10^5 N/C (C)</p> Signup and view all the answers

<p>1.08 $ 10^5 N/C (D)</p> Signup and view all the answers

<p>E1x + E2x (B)</p> Signup and view all the answers

<p>E = √(Ex^2 + Ey^2) (D)</p> Signup and view all the answers

<p>66.6° (B)</p> Signup and view all the answers

<p>It is used to find the force acting on it. (A)</p> Signup and view all the answers

<p>The combined electric fields E1 and E2. (C)</p> Signup and view all the answers

<p>An induced charge develops on the surface of the wall. (C)</p> Signup and view all the answers

<p>The charges are redistributed but stay within the sphere. (A)</p> Signup and view all the answers

<p>It obtains excess positive charge. (B)</p> Signup and view all the answers

<p>The separation distance and the product of their charges. (B)</p> Signup and view all the answers

<p>It is uncharged but has induced charges. (C)</p> Signup and view all the answers

<p>It means electric force weakens as distance increases. (A)</p> Signup and view all the answers

<p>The charges on the sphere's surface become uniformly distributed. (B)</p> Signup and view all the answers

<p>It can be polarized to show separation of charge. (A)</p> Signup and view all the answers

The force exerted by one charged object on another charged object is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The x-component of the force exerted by q1 on q3 can be calculated using Coulomb's Law and the distance between q1 and q3.

The x-component of the force exerted by q2 on q3 can be calculated using Coulomb's Law and the distance between q2 and q3.

The net force on q3 will be zero when the sum of the x-components of the forces exerted by q1 and q2 on q3 is equal to zero.

To find the unknown distance x, we can set the sum of the x-component forces equal to zero and solve the resulting quadratic equation.

The equation can be simplified by canceling out common factors like the Coulomb constant (k), the charges (q3, q1, q2), and the unit of 10^-6.

Re-arrange the simplified equation into the standard quadratic form, ax^2 + bx + c = 0.

The quadratic equation is used to solve for the distance (x) where the net force on q3 is zero.

A visual representation of electric fields using lines that indicate the direction and strength of the field.

The electric field vector at any point in space is tangent to the electric field lines at that point.

The density of electric field lines in a region is proportional to the strength of the electric field.

The electric field is stronger when the field lines are closer together and weaker when they are farther apart.

The electric field lines originate from positive charges and terminate on negative charges.

Electric field lines never intersect.

Electric field lines are continuous.

Electric field lines provide a visual representation of the direction and strength of the electric field, helping us understand the behavior of charged particles in that field.

The ability of an object to attract or repel other objects due to the presence of electric charges.

The process of redistributing charges within an object, usually by bringing a charged object nearby.

A material that allows electric charges to move freely through it. Examples include metals like copper and silver.

Material that resists the flow of electric charges. Examples include rubber, glass, and plastic.

The process of transferring charge from a charged object to another object by direct contact.

The process of transferring charge from a charged object to a neutral object without direct contact.

A material that can hold a charge for a long time, even after the charging source is removed.

The electric field strength at a point is defined as the force per unit charge experienced by a test charge placed at that point.

The magnitude of the electric force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The electric field strength at a distance from a point charge is inversely proportional to the square of the distance from the charge.

The electric field vector at a point is the vector sum of the electric field vectors due to all charges present.

The force exerted by a charge Q on a small test charge q0, divided by the test charge q0. It's a vector quantity describing the strength and direction of the electric field at a point.

A very small charge used to measure the electric field at a point. It's assumed to be so small that it doesn't significantly influence the field being measured.

The x-component of a vector is obtained by multiplying the magnitude of the vector by the cosine of the angle it makes with the x-axis.

The y-component of a vector is obtained by multiplying the magnitude of the vector by the sine of the angle it makes with the x-axis.

The SI unit for electric field strength. It represents one newton of force exerted on one coulomb of charge.

The magnitude of the resultant vector can be found using the Pythagorean theorem.

The electric field (E) is defined as the force (F) exerted on a test charge (q0) divided by the test charge (q0).

The test charge is assumed to be arbitrarily small to avoid interfering with the electric field being measured.

The direction of the resultant vector can be found using the inverse tangent function, which is the angle whose tangent is the ratio of the y-component to the x-component.

The electric field strength doesn't depend on the magnitude of the test charge, only on the charge creating the field.

The electric field at a point is the force per unit charge that a positive test charge would experience if placed at that point.

It's convenient to use a test charge of 1 coulomb (1C) for calculations, though it's not a physical test charge.

The process of finding the total electric field at a point by adding the individual contributions of multiple charges.

The electric field due to a charge extends radially outward from the charge if it is positive and radially inward if it is negative.

The magnitude of the electric field at a point is directly proportional to the magnitude of the charge causing the field and inversely proportional to the square of the distance from the charge.

The electric field at a point is the vector sum of the electric fields due to all charges present.

The x-component of the electric field is calculated using the cosine of the angle between the electric field vector and the x-axis, while the y-component is calculated using the sine of the angle.

To find the net electric field at a point, we add the x-components of each individual electric field and then add the y-components of each individual electric field.

The net electric field at a point due to multiple charges is the vector sum of the individual electric fields produced by each charge.