1804 United States presidential election in Rhode Island

Main article: 1804 United States presidential election

1804 United States presidential election in Rhode Island

November 2 - December 5, 1804

Nominee Thomas Jefferson Charles C. Pinckney (not on ballot) Party Democratic-Republican Federalist Home state Virginia South Carolina Running mate George Clinton Rufus King Electoral vote 4 0 Popular vote 1,312 — Percentage 100.00% —

President before election Thomas Jefferson Democratic-Republican Elected President Thomas Jefferson Democratic-Republican

The 1804 United States presidential election in Rhode Island took place as part of the 1804 United States presidential election. Voters chose 4 representatives, or electors to the Electoral College who voted for president and vice president.

Rhode Island voted for the Democratic-Republican candidate, Thomas Jefferson, over the Federalist candidate, Charles C. Pinckney. Jefferson won Rhode Island by a margin of 100.00%, the largest margin and landslide victory ever since in the state. Because Pinckney was not on the ballot.

Results

1804 United States presidential election in Rhode Island
Party		Candidate	Votes	Percentage	Electoral votes
	Democratic-Republican	Thomas Jefferson (incumbent)	1,312	100.00%	4
	Federalist	Charles C. Pinckney (not on ballot)	—	—	0
Totals			1,312	100.0%	4

See also

• United States presidential elections in Rhode Island